THEPOWEROFVEDICMATHS

2ndEDITION

ATULGUPTA

PublishedbyJaicoPublishingHouse

A-2JashChambers,7-ASirPhirozshahMehta

RoadFort,Mumbai-400001jaicopub@jaicobooks.comwww.jaicobooks.com

©AtulGupta

THEPOWEROFVEDICMATHS

ISBN81-7992-357-6

FirstJaicoImpression:2004

ThirteenthJaicoImpression(Revised&

Updated):2010FifteenthJaicoImpression:2011

Nopartofthisbookmaybereproducedorutilizedinanyformorbyany

means,electronicormechanicalincluding

photocopying,recordingorbyanyinformationstorageandretrievalsystem,without

permissioninwritingfromthepublishers.

PREFACE

Mathematics isconsideredtobeadryandboring subject by a largenumber of people.Children dislike and fearmathematics for a varietyof reasons. This book iswritten with the sole

purposeofhelpingschooland college students,teachers, parents,common people andpeople from non-mathematical areas ofstudy,todiscoverthejoysof solving mathematicalproblems by a wonderfulset of techniques called‘VedicMaths’.

These techniques are

derived from 16 sutras(verses) in the Vedas,which are thousands ofyears old and among theearliest literature ofancient Hindus in India.They are an endlesssource of knowledge andwisdom, providingpracticalknowledgeinallspheresof life. JagadguruSwamiSriBharatiKrshna

Tirthaji (1884-1960) wasa brilliant scholar whodiscovered the 16 sutrasin the Vedas and spent 8years in their intensestudy. He has left aninvaluable treasure for allgenerations to come,consisting of a set ofunique and magnificentmethods for solvingmathematical problems in

areas like arithmetic,algebra, calculus,trigonometry and co-ordinate geometry. Thesetechniques are very easyto learn and encapsulatethe immense and brilliantmathematical knowledgeof ancient Indians, whohad made fundamentalcontributions tomathematics in the form

of the decimal numerals,zeroandinfinity.

I have trainedthousands of children ofall age groups with thesetechniquesand I find thatevenyoungchildrenenjoylearning and using them.The techniques reducedrastically, thenumberofsteps required to solveproblems and in many

cases, after a littlepractice, many of theproblems can be solvedorally. It givestremendous self-confidencetothestudentswhichleadsthemtoenjoymathematics instead offearinganddislikingit.

I have written thisbook in the form of acookbook,wherea reader

can grasp a techniquequickly, instead ofreading through a largemass of theory beforeunderstanding it. I haveconsidered techniques formajor arithmeticaloperations likemultiplication, division,computation of squaresand square roots andcomplexfractions,besides

a whole lot of othertechniques. Eachtechnique has beenexplained in detail withthe help of solvedexamples, using a step-by-stepapproachandIamsurethatthereaderwillbeable to understand andmasterthecontentseasily.Every chapter has a largenumber of problems for

practice and the bookcontains over 1000 suchproblems. The answersare given alongside sothat the reader can eithersolve the problems orallyor use paper and pen andcompare with the givenanswer. The chaptersshould be readsequentially to absorb thematerial and then can be

used for reference in anydesiredorder.

Ihavealso includedaspecialchapterinwhichIhave shown theapplication of thetechniques to solveproblems, collected fromseveral competitiveexams. This is a uniquefeature of the book andshould add to the

popularity of thetechniques.

I have tried to makeall the examples andanswers error-free but ifanymistakeisdiscovered,I will be obliged if I aminformedaboutthesame.

Constructive criticismandcommentscanbesentto me at

greglobal@gmail.com

ATULGUPTAMobile:9820845036

Contents

Preface

Chapter 1 Two SimpleTechniques

Subtraction from100/1000/10000NormalMethod

VedicMethod

Multiplicationwithaseriesof9s

Multiplication of anumber by samenumberof9s

Multiplication of anumber by greaternumberof9s

Multiplication of anumber by lesser

numberof9s

Chapter 2 Operationswith9

Computation ofremainder(Navasesh) ondivisionby9

Basicmethod

Firstenhancement

Secondenhancement

Final compactmethod

Verification of theproduct of twonumbers

Verification of thesumoftwonumbers

Verification of thedifference of two

numbers

Verification of thesquare or cube oftwonumbers

Limitation in theverificationprocess

Computation of thequotient on divisionby9

Method1

Method2

Chapter 3 Operationswith11

Multiplication

Divisibility test ofnumbersby11

Multiplication with111

Chapter 4

Multiplication(Nikhilam)

Secondary bases of50

Secondary base of250

Secondary base of500

Secondarybases like40,60

Secondary base of300

Chapter 5Multiplication (UrdhvaTiryak)

2-Digitmultiplication

3-Digitmultiplication

Multiplying 3-digitand2-digitnumbers

4-Digitmultiplication

Multiplying 4-digitand3-digitnumbers

Chapter6Division

Divisionbyaflagofone digit (no

remainder)

Divisionbyaflagofone digit (withremainder)

Division withadjustments

Division with a flagof2digits

Division with a flagof3digits

Chapter 7 SimpleSquares

Chapter 8 Square ofAnyNumber

Definition -Dwandwa orDuplexSquare of anynumber

Square of anynumber

Chapter 9 Square RootofaNumber

Chapter 10 Cubes andCubeRoots

Computing cubes of2-digitnumbers

Cuberootsof2-digitnumbers

Computing fourth

power of 2-digitnumbers

Chapter 11Trigonometry

Triplet

Computingtrigonometricratios

Computingtrigonometric ratios

oftwicetheangle

Computingtrigonomegtric ratiosofhalftheangle

Chapter 12 AuxiliaryFractions

Divisors endingwith‘9’

Divisors endingwith

‘1’

Divisors endingwith‘8’

Divisors endingwith‘7’

Numbers endingwith‘6’

Otherdivisors

Chapter13Mishrankor

Vinculum

Conversion toMishrank

Conversion fromMishrank

Application inaddition

Application insubtraction

Application inmultiplication

Application indivision

Application insquares

Applicationincubes

Chapter 14SimultaneousEquations

Chapter15Osculator

Positiveosculators

Negativeosculators

Chapter16ApplicationsofVedicMaths

Sampleproblems

Solutions usingVedicmaths

Problemsforpractice

Answers

TWOSIMPLETECHNIQUES

We will begin our

journey into thefascinating world ofVedic maths with twosimple techniques whichwilllaythefoundationforsomeof the techniques inthefollowingchapters.

I. Subtraction from100/1000/10000

We will start with avery simple technique

wherein we will see theuseof thesutra ‘All from9 and last from 10’. Thisisusedtosubtractagivennumber from 100, 1000,10000 etc. It removes themental strain which isexistent in the methodtaughtinschools.

This method is alsoused later on in theNikhilam method of

multiplication.

Consider thesubtraction of 7672 from10000.

a)NormalMethod

Thenormalmethodis

Wecarry‘1’fromthe

left side and continuedoingso tillwe reach therightmost digit, leavingbehind 9 in each columnand10inthelastcolumn.

Then, we subtract theright most digit ‘2’ from10 and write down ‘8’.Next,wesubtractthedigit‘7’ from ‘9’ and writedown‘2’.

Werepeatthisprocessfor all the remainingdigitstotheleft.

Through thisoperation, the final resultis always obtained fromrighttoleft.

Mentally, there is acarry operation for everydigit, which is timeconsuming and slows

downtheoverallprocess.

b)VedicMethod

The Vedic methodusesthesutra‘Allfrom9and last from 10’ andgives a very simple andpowerful technique toachievethesameresult.

The result can beobtainedfrombothleft toright as well as right to

left with equal ease. Itstates that the result canbeobtainedbysubtractionofeachdigitfrom‘9’andthelastdigitfrom‘10’.

Hence, in the givenexample,

We can get the resultfrom left to right or vice

versafromrighttoleftas

i.e. all digits except thelast one are subtractedfrom 9, the last digit issubtracted from 10 andtheresult(2328)iswrittendowndirectly.Thementalburdenofacarryforeachcolumn vanishes and theanswer can be obtained

easily,inajiffy.

The same techniquecanbeappliedfordecimalsubtraction also, e.g.2.000-0.3436.

The core operationhere is subtraction of3436from10000where1isacarryfromleft.

1. Examples for

subtractions from basesof100,1000etc.

This simplesubtraction method willbe used further inmultiplication (Ch.4) andmishrank(Ch.13).

II. Multiplicationwithaseriesof9s

a)Multiplicationofanumber by samenumberof9s

Let us use thistechnique to see how tocarryoutmultiplicationofan n-digit number by anumber with ‘n’ numberof 9s eg. any 3-digitnumber by 999 or a 4-digitnumberby9999.

Letusseeanexampleviz.533×999

The solution can be

obtainedas533×(1000-1)

= 533000 -533

=532467.

Theanswerconsistsoftwo 3-part numbers viz.‘532’and‘467’.

The first part ‘532’ is1 less than the given

number,i.e.533-1.

The secondpart ‘467’isequalto(1000-533)or(999 - 532). It is simplythe ‘9’s complement ofthe first 3 digits of theresult; i.e. 9s complementof532wheretherequireddigitscanbeobtainedas:

4=9-5

6=9-3

7=9-2

Similarly, 3456 ×9999 will consist of twoparts

3455/6544

i.e. 34556544 (where,6544 is ‘9’s complementof3455).

This simple techniquecan be used to get the

resultorallywheneveranynumberismultipliedbyanumber consisting of anequalnumberof9s.

The same techniquecan be used formultiplication of decimalnumbers.

Consider the example4.36×99.9

Firstly, multiply 436

by 999 and write downtheresultas435564.

Now, since thenumbers have two andone decimal placesrespectively, the finalresult will have threedecimalplacesandweputthe decimal point threepoints from the right. So,the final answer will be435.564.

b)Multiplicationofanumber by greaternumberof‘9’s

Letusnowseehowtocarryoutmultiplicationofan n-digit number by anumber with greaternumberof‘9’s

E.g. 235 by 9999 or235by99999

i) Consider 235 by9999

Since the number ofdigitsin235isthreewhilethere are four digits in9999, we pad 235 withone zero to get 0235 andcarry out themultiplicationasbefore.

0235 × 9999 =0234/9765 and the final

resultis2349765.

ii) Consider 235 by99999

Padding235with twozeroes,weget

00235 × 99999 =00234/99765andthefinalresultis23499765.

c)Multiplicationofanumber by lesser

numberof9s

Letusnowseehowtocarryoutmultiplicationofan n-digit number by anumber with lessernumberof9s.

i)Consider235by99

Since there are twodigits in 99, we place acolon twodigits from the

right in thegivennumbertoget

2:35

We enclose thisnumberwithinacolonontherighttoget

2:35:

We now increase thenumber to the left of thecolon by 1 and derive a

newnumberas

3:35

Wealign thecolonofthe new number with theenclosing colon of theoriginalnumbertoget

2:35:3:35

We now subtract 3from 235 and 35 from

100, to get the finalansweras

ii)Consider12456by99

Since there are twodigits in 99, we place acolon twodigits from theright in thegivennumber

to get 124 : 56 Weenclose this numberwithinacolonontherighttoget

124:56:

We now increase thenumber to the left of thecolon by 1 and derive anewnumberas

125:56

Wealign thecolonofthe new number with theenclosing colon of theoriginalnumbertoget

124:56:

125:56

We now subtract togetthefinalansweras

2. Examples formultiplication by seriesof9s

In examples where

decimal numbers areinvolved,wemultiply thetwo numbers assumingthat therearenodecimalsand then place thedecimalintherightplace,asexplainedabove.

OPERATIONSWITH9

The number 9 has a

special significance inVedic maths. We willstart by looking at asimple but powerfultechniquewhichisusedtoget the remainder ondividinganynumberby9.This remainder is givenseveral names like‘Navasesh’, ‘Beejank’ ordigital root. It haspowerfulpropertieswhich

are used to check thecorrectness of arithmeticoperations like addition,subtraction,multiplication, squaring,cubing etc. It can also beused to check thedivisibility of any givennumberby9.

We will begin byseeing how to computethenavaseshofanygiven

number and then proceedtoseeitsapplications.

I. Computation ofremainder (Navasesh)ondivisionby9

The navasesh of anumber is defined as theremainder obtained ondividingitby9.

Example : The

navaseshof20is2Thenavasesh of181is1Thenavasesh of8132is5.Thenavasesh of2357is8

Letus seehow tousetechniques provided by

Vedic maths to computethe navasesh for anynumber.Wewilldenoteitby‘N’.

a)BasicMethod

To compute thenavasesh (N) add all thedigitsinthenumber.

If we get a singledigit as the answer,

then that is thenavasesh (if the sumis 0 or 9, then thenavaseshis0).If there ismore thanonedigit in the sum,repeattheprocessonthe sum obtained.Keep repeating till asingle digit isobtained.

Examples:

Navaseshof20canbecomputed by adding thedigits 2 and 0 to get theresultas2.

Navasesh of 8132 issimilarly obtained asfollows:

Pass1:8+1+3+2gives14,

which has more thanonedigit.

Pass2:1+4gives5which is the required

navasesh.

b) FirstEnhancement

Ifthereisoneormore‘9’intheoriginalnumber,these need not be addedand can be ignoredwhilecomputing thesumof thedigits

Example : If thenumberis23592,then‘N’can be computed asfollows.

Pass1:2+3+5+2=12(ignore9)

Pass2:1+2=3.Sothenavaseshis3.

c) SecondEnhancement

Whilesummingupthedigits, ignore groupswhichaddupto9.

Example : If thenumberis23579,then‘N’can be computed asfollows.

Ignore ‘9’ asexplainedabove

Ignore 2 + 7 (ie 1stand 4th digits which add

upto‘9’)andaddonly‘3’and‘5’togetthe‘N’as8.

d) Final CompactMethod

As before, ignore all‘9’sandgroupswhichaddupto9.

Whilesumminguptheremaining digits, if thesumaddsuptomorethan

two digits at anyintermediatepoint,reduceit to one digit there itselfby adding up the twodigits.Thiswill eliminatethe need for the secondpass which is shown intheexamplesabove.

Example : If thenumberis44629,then‘N’can be computed asfollows.

Ignore ‘9’ in the lastposition.

Add the remainingdigits from the leftwherethefirstthreedigits,viz4+4+6addupto14.

Since, this has twodigits,weaddthesedigitshere itself to get a singledigiti.e.1+4=5

Now add ‘5’ to the

remaining digit ‘2’ to getthe‘N’as7.

All methods give thesame answer, but theresult can be obtainedfaster and with lessercomputationwhenweusetheimprovedmethods.

II)Verificationoftheproductoftwonumbers

Thecorrectnessofanyarithmetic operation canbe verified by carryingoutthesameoperationonthe navasesh of thenumbers in the operation.Oncewehave learnthowto compute the navaseshofanumber,wecanuseitto check whether theresult of operations likemultiplication, addition

andsubtractionontwo(ormore) numbers is correctornot.

Letusseeanexample.

Let’s take the productof38×53whichis2014.

Howdoweverify thecorrectness of the answer‘2014’?

Let’s take the ‘N’ of

each of the multiplicandsandoftheproductandseethe relation between thesame

Now, N(38) = 2 ..Navasesh of the firstnumber

N(53) = 8 ..Navaseshof thesecondnumber

N(2014) = 7 ..

Navaseshof theproduct

Consider the productof the ‘N’ of the 2multiplicandsi.e.N(38)×N(53)=2×8=16whosenavaseshinturnis7.

The‘N’oftheproductisalso‘7’!!

Thus the product ofthe navasesh of the two

multiplicands is equal tothe navasesh of theproductitself.Thissimpletest can be used to verifywhether the answer iscorrect.

AnotherExample

22.4×1.81=40.544

On ignoring thedecimal points, thenavasesh of the two

numbersare

N(224)=8andN(181) = 1 and the

productofthenavaseshis8.

The navasesh of theproduct is also N(40544)=8andhence theanswerisverified.Thepositionofthe decimal point can beverifiedeasily.

III) Verification ofthesumoftwonumbers

Similarly,let’sseethecaseofthesumofthetwonumbers38+53whichis91.

We see that N(38) +N(53)=2+8=10whichin turn has a navasesh of1.

Therefore the

navaseshofthesumofthenavaseshis1.

The ‘N’ of thesum(91)isalso1!!

Thus the sum of thenavasesh of the twonumbers is equal to thenavaseshofthesumitself.

IV) Verification ofthe difference of two

numbers

Let’s see the case ofthe difference of the twonumbers.

53-38=15N(53)–N(38)=8-2=6N(15)=6

Thusthedifferenceofthe navasesh of the twonumbers is equal to the

navaseshofthedifferenceitself.

Sometimes, thedifferenceofthenavaseshmaybenegative.Then, itcan be converted to apositivevaluebyadding9toit.

ThusN(–5)=4,N(–6)=3.

Let us take another

example,viz

65-23=42N(65)=2N(23)=5N(65)–N(23)=–3,

As explained above,N(–3)=6

The navasesh of theresultshouldbe6.

We see that thenavasesh of 42 is indeed6.

Hence,wecanusethismethod to verify thedifference of twonumbers.

V)Verificationofthesquare or cube of twonumbers

Let’s take the case ofsquaring a number, say,34.

342=1156

Now take thenavasesh on both sides,i.e. navasesh of 34 is 7andnavaseshof1156is4.

Let us square thenavasesh of the left side

i.e. 72 = 49 whosenavaseshisalso4.

Thusthesquareofthenavasesh of the givennumber is equal to thenavasesh of the square.The reader can tryverifying for the cube ofanynumberbyhimself.

VI)Limitationintheverificationprocess

This verificationmethod has a limitationwhich has to be kept inmind.

Consider again theexampleof

38×53=2014

WeknowthatN(38)×N(53) should equalN(2014)

On the left side, we

have2×8=16which,inturn, gives a navasesh of7.

Now,iftheproductonthe RHSwas assigned as2015, its ‘N’ would be 8from which we wouldcorrectlyconcludethattheresultiswrong.

But if theanswerwaswrittenas2041insteadof

2014, the ‘N’ would stillbe 7, leading us wronglyto the conclusion that theproductiscorrect.

So, we have toremember that thisverification process usingnavasesh,wouldpointoutwrongly computed resultswith 100% accuracy butthere can be more thanone result which would

passthe‘correctness’test.We have to be carefulaboutthis.

In spite of thislimitation, it remains averypowerfulmethod forverification of resultsobtained by any of thethree main arithmeticoperations.

VII)Computation of

the quotient on divisionby9

Letusnowseehowtoget the quotient ondividing a number by 9.In Vedic maths, thedivision is converted to asimpleadditionoperation.

a)Method1

The first method

consistsofaforwardpassfollowed by a backwardpass.

Example : Divide8132by9

Steps:

Place a colon beforethe last digit, which

shows the positionseparating thequotient and theremainder.

Start from the leftmost digit and bringitdownasitis.

Carry it (8) to thenextcolumnandaddit to the digit in thatcolumni.e.1+8and

bringdown9.

Carry 9 to the nextcolumn, add it to 3andbringdown12.

Repeat by carrying12 to the next (last)column,addto2andbring down 14 asshown.

After reaching the

end, we workbackwards from theright to the left bycomputing theremainder in the lastcolumn and bycarrying the surplusdigits to the left.Wehave to retain onlyone digit at eachlocation.

Examine the last

column, whichshould hold theremainder. If thenumber ismore than9, we subtract themaximum multiplesof 9 from it, whichgives the quotientdigit to be carried tothe left, leaving theremainderbehind.

Since the number in

the last column (14)containsonemultipleof 9, we carry 1 tothe left hand digit(12)andretain5 (14- 9) as finalremainder.

At the end of theforward pass, thedigits looked asfollows:

8.9.12:14.

whichnowbecome

8.9.13:5.

Now retain 3 andcarry 1 to the left toget10.

8.10.3:5.

Next, retain 0 and

carry 1 to the left toget9.

9.0.3:5.

So the finalcomputation wouldappearasfollow

The quotient is 903and the remainder is 5.Notice that the steps areverysimpleandverylittlemental effort is requiredtoobtaintheanswer.

b)Method2

Here,wewillcomputethe result in the forwardpass itself. Duringcolumn-wise

computation, if the digitinanycolumnbecomes9or more, we can transferthe multiples of 9 to theleft and retain theremainderinthecolumn.

In theaboveexample,at the second column,whenweget 9,we retainzero(9-9)inthecolumnand carry one to the left.Wethentake0tothenext

column.

Here,wewillsubtract9 from 9 to retain 0 andcarry 1 to the left. Onadding1to8,wewillget9 and the computationwilllookasshownbelow.

This eliminates theneed for the backwardpass and increases theoverall speed ofcomputation.

AnotherExample

Consider the divisionof8653by9

So,thequotientis961andtheremainderis4.

3. Examples fornavasesh

OPERATIONSWITH11

Inthischapterwewillconsider two distinctoperations with thenumber 11. Firstly, wewill see how to multiplyanynumberofanylengthby11andwritetheresult,orally in a single line.Secondly, we will seehowtotestthedivisibilityofagivennumberby11,i.e. whether a given

number is fully divisibleby 11 or not. These areuseful techniques and ifwhile solving a problem,one of the intermediatesteps leads tomultiplicationby11oritsmultiple, this techniquecan be used to solve theproblemquickly.

Oncethetechniqueformultiplication by 11 is

mastered, multiplicationofanumberby22,33,44etc, amultiple of 11, canbe carried out quickly bysplitting the multiplicandas11×2,11×3or11×4. The given number canbe multiplied by 11 andthen the product can bemultiplied further by theremaining digit, i.e. 2 or3,asthecasemaybe.

Multiplication

I.Letusconsiderthemultiplication of 153 by11.

The traditionalmethod is carried out asfollows:

The faster, one linemethodisasfollows:

We obtain each digitof the result by addingpairs of numbers asexplainedbelow:

Steps

Start from the right

and write down thedigit3directly

Form pairs ofconsecutive digits,starting from therightandwritedownthe sum, one at atime.

Start from the right,form the first pairconsistingof3and5,

add them and writedown the result, i.e.writedown8

Move to the left,formthenextpairbydropping the right-most digit 3 andincluding the nextdigiti.e.1.

Add 5 and 1 andwritedowntheresult

6.

Now,ifwemoveleftand drop the digit 5,weareleftonlywithonedigit,i.e.theleft-mostdigit.

Write down the leftdigit1directly.

Thus the final answeris 1683. Isn’t that real

fast?

II. Let us nowmultiply25345by11

The result is obtainedstep by step as shownbelow:

Steps

This gives the resultof25345×11=278795

III.Let’stryanotherexample:157×11

Wewill use the sametechniqueasfollows:

Here,theadditionof7and 5 gives 12, whereinweretain2andcarry1tothe left. The next pairconsistsof5and1whichon addition gives 6. Addthe carry digit 1 to it toget7.

So the final answer is1727, which can beverified by the traditionalmethod.

Ifthemultiplicationisbetween decimalnumbers,wecarryoutthemultiplication as if thereare no decimals and thensimply place the decimalpointattherightplace.

4. Examples formultiplicationby11

Divisibility test ofnumbersby11

How do we testwhether a number isdivisibleby11?

Example : Is 23485divisibleby11?

Steps

Start from the right

(unit’s place) andadd all the alternatenumbers i.e. all thedigits in the oddpositions viz. 5, 4and 2 which gives11.

Now, start from thedigitinthetensplaceand add all thealternatenumbersi.e.all the digits in the

even positions viz. 8and 3 which gives11.

Ifthedifferenceofthetwosumsthusobtainediseitherzeroordivisibleby11,thentheentirenumberisdivisibleby11.

In theexampleabove,thedifference is zero andhence the number is

divisible.

Let’s take anotherexample and check thedivisibilityof17953.

Thesumofthedigitsin the odd positionsis3+9+1whichisequalto13.

Thesumofthedigitsin the even positions

is 5 + 7 which isequalto12.

Sincethedifferenceofthesetwosumsis1whichis neither 0 nor divisibleby 11, the given numberisnotdivisibleby11.

Ifthegivennumberis17952 instead of 17953,thetwosumsare12(2+9+ 1) and 12 (5 + 7). The

difference of the twosums is zero and hencethe given number isdivisibleby11.

Exercises

1.Is8084divisibleby11?Ans(No)

2. IfX381 isdivisibleby11,whatisthesmallestvalueofXAns(7)

Multiplication with111

We can use the sametechnique to multiply anumber by 111. As with11, we proceed with thecomputation from theright starting with 1 digitand progressivelyincreasing the digits tillwe reach 3 digits. Anexample will make it

clear.

E.g.4132×111

The result isobtainedstepbystep as shownbelow:

Steps

This gives the resultof4132×111=458652

The technique can beextendedfornumberslike1111,11111etc.

MULTIPLICATION(NIKHILAM)

Multiplication of

numbers plays a veryimportant role in mostcomputations. Vedicmaths offers two mainapproaches tomultiplying2numbers.Thesearethe

Nikhilam methodandUrdhva Tiryakmethod

In the Nikhilam

method we subtract acommon number, calledthe base, from both thenumbers. Themultiplication is thenbetweenthedifferencessoobtained.

Inthenextchapterwewill take up the ‘UrdhyaTiryak’ method, whichusescross-multiplication.

Let us start the studyof the Nikhilam methodnow by consideringmultiplicationof numberswhich are close to a baseof 100, 1000, 10000 etc.Later on, wewill see themultiplicationof numbersclose to bases like 50,500,5000etc.andfinallyanyconvenientbase.

For a base 100, the

numbers which are beingmultipliedmaybe

less than the base(e.g.96,98)more than the base(e.g.103,105)mixed (e.g. 96 &103)

Case1:Integerslessthanthebase(100)

Consider themultiplicationof93×98.Let’s look first at thetraditional method andthenatVedicmethod.

VedicMethod

Write the twonumbers one below theother and write thedifference of the base(100) from each of thenumber, on the right sideasshown:

Take the algebraicsum of digits across anycrosswise-pair.e.g.93+

(-2)or98+(-7).

Both give the sameresultviz.91,whichisthefirstpartoftheanswer.

This is interpreted as91 hundreds or 9100, asthebaseis100.

Now,justmultiplythetwo differences (–2) × (–7) toget14,which is thesecondpartoftheanswer.

The final answer isthen9114,whichcanalsobe interpreted as 9100 +14.

ImportantPoints

Since thebase is 100,weshouldhavetwodigits(equal to number ofzeroes in the base) in thesecondpartoftheanswer.

If the number ofdigits is one, thenumber is paddedwithonezeroon theleft.

If the number ofdigits is more thantwo, then the leftmost digit (in the

hundred’s position)iscarriedtotheleft.

The method isequally effectiveeven if one numberis close to the baseandtheotherisnot.

Case 2 : Integersabovethebase(100)

Consider themultiplication of 103 ×105. The techniqueremains exactly the sameand can be written asfollows:

Notice that here thedifferences are positive,i.e.+3(103-100)and+5(105-100).

The addition in thecross-wisedirectiongives108(103+5or105+3).

The product of thedifferences is 15 and sothefinalresultis10815.

Once again, we have

to ensure that we haveexactly two digits in thesecondpartoftheanswer,which is achieved byeitherpaddingwithzeroesorbycarrytotheleft.

E.g. Consider thefollowingtwocases

Here, 13 and 15 havebeen multiplied by usingthesametechnique,wherethebasehasbeentakenas10.

This same techniquecan also be used to find

the squares of numbersclose to thebase,e.g.962

or1042.

Case 3 : Integersbelow and above thebase(100)

Consider themultiplication of 97 ×104.

Here 97 is below 100

while104isabove100.Using the same

technique,weget

Here the 2nd part ofthe answer is -12. Wenowcarry1 from the leftwhichisequivalentto100(sincethebaseis100)and

the final answer isobtainedasfollows:

101 / (–12) = 100 /(100-12)

= 100 / 88=10088.

This can also beinterpreted as 10100 - 12=10088.

Use the technique

explainedinchapter1(allfrom 9 and last from 10)tosubtractfrom100.

Once again, the 2ndpartofthenumbershouldhave exactly two digits,since the base has 2zeroes.

Consider thefollowing2examples:

132 can be obtainedby multiplying 12 by 11asexplainedinChapter2.Since 132 is more than100, we have to carry 2

from the left, which willbe equivalent to200 (2×100).

Hence the resultwouldbe

= 99 / (200 - 132) =9968

Or as before, we canget the answer byinterpretingitas

10100-132=9968.

Case4:Examplesofdecimalnumbers

Consider9.8×89.Wewill assume the numberstobe98and89andcarryout the multiplication asshownbelow:

Now, since theoriginalnumbershaveonedigitinthedecimalplace,we put the decimal pointone digit from the rightandgetthefinalansweras872.2.

Case 5 : Exampleswith bases of 1000 and10000

Consider the example997×993,wherethebaseis1000.

The procedureremainsthesame.

The first part of theanswer i.e. 990 isobtainedbythecross-wisesum,993-3or997–7.

The second part isobtainedbymultiplying3and7andpadding21with‘0’ since we should have3digits.

Hence the finalansweris990021.

Now, consider theexample 9998 × 10003,wherethebaseis10000.

The first part of theanswer i.e. 10001 isobtainedbythecross-wisesum,9998+3or10003-2.

The second part isobtainedbymultiplying–2 and +3 and padding –6with ‘000’ since weshould have 4 digits(numberofzeroesinbase10000). We now carry 1from the left, (equivalentto 10000) and subtract 6fromittoget9994.

As before, this is100010000-6.

Hence the finalanswer is 10000/9994 i.e.100009994.

The method formultiplying decimalnumbers has beenexplained above whereinwe first consider thenumbers without decimalpoints, multiply themusing this technique, andthen place the decimal

point at the appropriatelocation.

Case6:Otherbases—50,250,500etc

Let us now considerthe multiplication ofnumbers which may beclose to other bases like50, 250, 500 etc. whicharemultiplesof50.Later,we will see how the

method can be extendedtobaseslike40,whicharenot multiples of 50. Thereaderwillthenbeabletoextend themethod to anybaseofhischoice.

The solved examplesaregroupedunder:

i. Secondary base of50

ii. Secondary base of

250iii. Secondary base of

500iv. Secondary base of

40and60v. Secondary base of

300

I. Secondary base of50

Example 1 -Consider the

multiplicationof42×46

Here,we can take thesecondarybaseas50.Thecorresponding primarybasecanbeconsideredas10 or 100. The scalingfactor between theprimary and secondarybasewouldbeasfollows:

Factor = Secondarybase/Primarybase

5=50/10(ifprimarybase=10)

1/2 = 50 / 100 (ifprimarybase=100)

We can use either ofthese scaling operationsand compute the requiredanswer. Let’s see eachcaseindetail.

A)Case 1 - Primary

base 10 and secondarybase50for42×46

Secondarybaseis50Scaling operation :Factor=5=50/10

(secondarybase ÷primarybase)

One digit to beretained in the righthand part of theanswer because

primarybasehasonezero.

The initial part of themethod remains the sameandisasfollows:

Consider : 42 × 46(subtract 50 from eachnumber)

Now,wehavetoscaleup the left part of thisintermediate result bymultiplyingby5.

Since 38 × 5 = 190,theanswerbecomes190/32.

Sincetheprimarybase(10)containsonezero,wewillretainonedigitintheright part of the answer

andcarry3totheleft.

So, the final answerbecomes1932.

B)Case 2 - Primarybase 100 and secondarybase50for42×46

Secondarybaseis50Scaling operation :Factor = 1/2 = 50 /100

(secondarybase ÷primarybase)

Two digits to beretained in the righthand part of theanswer.

The techniqueremainsthesameandisasfollows:

Consider : 42 × 46

(subtract 50 from eachnumber)

Now,wehavetoscaledown the left part of thisintermediate result bymultiplyingitby1/2.

Since38 /2=19, theanswerbecomes19/32=

1932.

Sincetheprimarybaseis 100, which containstwozeroes,wewillretaintwodigitsintherightpartoftheanswer.

So, the final answerbecomes1932.

Wow, isn’t thatwonderful? Both theoperations give us the

sameresult.

SpecificSituations

Letusnowconsiderafewspecificsituations:

A fraction isobtained during thescaling operationand/orThe right part isnegative

We will take anexample having boththesesituations.

Example 2 -Consider themultiplicationof52×47

C)Case 1 - Primarybase 10, secondary base50for52×47

Secondarybaseis50

Scaling operation :Factor=5=50/10

(secondarybase ÷primarybase)

One digit to beretained in the righthand part of theanswer

Consider : 52 × 47(subtract 50 from eachnumber)

We will scale up thisintermediate result bymultiplying49by5.

Since 49 × 5 = 245,theanswerbecomes245 /(–6)

Sincetheprimarybaseis 10, we will carry one

from the left which isequalto10andsubtract6fromit.

So the final answerbecomes244/4=2444.

This can also beconsidered as 2450 - 6 =2444.

D)Case 2 - Primarybase 100 and secondary

base50for52×047

Secondarybaseis50Scaling Operation :Factor = 1/2 = 50 /100

(secondarybase ÷primarybase)

Two digits to beretained in the righthand part of theanswer

Consider : 52 × 47(Subtract 50 from eachnumber)

We will scale downthisintermediateresultbydividing49by2.

Since 49 / 2 = 24.5,theanswerbecomes24½/

(–6)

Sincetheprimarybaseis 100, we will carry 1/2fromtheleft.Thisisequalto a carry of 50 (primarybase 100/2). We willsubtract6 from50giving44.

So the final answerbecomes24/44=2444.

Since24.5hundredsis

2450, this can beinterpreted also as 2450 /(–6)=2444

We get the sameanswer irrespectiveof theprimary base. Thefractional part can beavoided if we choose ascaling operationinvolving multiplicationinsteadofdivision.

II.Secondarybaseof250

Example 2 -Consider themultiplication of 266 ×235

E) Primary base1000andsecondarybase250for235×266

Secondary base is

250Scaling operation :Factor= 1/4= 250 /1000

(secondarybase ÷primarybase)

Three digits to beretained in the righthand part of theanswer

Consider:235×266

(subtract 250 from eachnumber)

Considerthesteps

We will scale downthisintermediateresultbydividing251by4.

Since 251 / 4 = 62¾,

theanswerbecomes62¾/(–240).

Sincetheprimarybaseis1000,wewillcarry3/4from the left which isequal to a carry of 750(primary base 1000 * ¾)andsubtract240from750giving510.

So the final answerbecomes62/510=62510.

Or, it can beconsideredas62750-240=62510.

III. Secondary baseof500

Example 2 -Consider themultiplication of 532 ×485

F) Primary base 100

and secondary base 500for532×485

Secondary base is500Scaling operation :Factor = 5 = 500 /100

(secondarybase ÷primarybase)

Two digits to beretained in the right

hand part of theanswer

Consider:532×485(Subtract500fromeachnumber)

Considerthesteps

We will scale up this

intermediate result bymultiplying517by5.

Since517×5=2585

So the answerbecomes2585/(–480).

Sincetheprimarybaseis 100, we will carry 5from the left which isequal to a carry of 500(primary base 100 * 5)andsubtract480from500

giving20.

So the final answerbecomes 2580/020 =258020.

Or, it can beconsidered as 258500 -480giving258020.

G) Primary base1000andsecondarybase500for532×485

Secondary base is500Scaling operation :Factor= 1/2= 500 /1000

(secondarybase ÷primarybase)

Three digits to beretained in the righthand part of theanswer

Considerthesteps

We will scale downthisintermediateresultbydividing517by2.

Since517/2=258½,theanswerbecomes258½/(–480).

Sincetheprimarybaseis 1000, we will carry ½from the left which isequal to a carry of 500(primary base 1000 * ½)andsubtract480from500giving20.

So the final answerbecomes 258/020 =258020.

The left part of the

answer is 258.5 hundredswhichisequalto258500.

So, it can beconsidered as 258500 -480giving258020.

IV. Secondary baseslike40,60

Example 3 -Consider themultiplicationof32×48

H) Primary base 10and secondary base 40for32×48

Secondarybaseis40Scaling operation :Factor=4=40/10

(secondarybase ÷primarybase)

One digit to beretained in the righthand part of the

answer

Consider : 32 × 48(subtract 40 from eachnumber)

We will scale up thisintermediate result bymultiplying40by4.

Since 40 × 4 = 160,theanswerbecomes160/(–64).

Sincetheprimarybaseis10,wewillcarry7fromtheleftwhichisequaltoacarryof70 (primarybase10 * 7) and subtract 64from70giving6.

Thisisalsoequivalentto1600-64.

So,thefinalansweris1536.

Example 4 -Consider themultiplicationof64×57

I) Primary base 10and secondary base 60for64×57

Secondarybase=60Scaling operation :

Factor=6=60/10(secondarybase ÷primarybase)

One digit to beretained in the righthand part of theanswer

Consider : 64 × 57(subtract 60 from eachnumber)

We will scale up thisintermediate result bymultiplying61by6.

Since 61 × 6 = 366,theanswerbecomes366/(–12)=364/8.

Sincetheprimarybaseis10,wewillcarry2from

theleftwhichisequaltoacarryof20 (primarybase10 * 2) and subtract 12from20giving8.

Thisisalsoequivalentto3660-12.

So,thefinalansweris3648.

V.Secondarybaseof300

Example - Considerthemultiplicationof285×315

J) Primary base 100and secondary base 300for285×315

Secondary base =300Scaling operation :Factor = 3 = 300 /100

(secondarybase ÷primarybase)

Two digits to beretained in the righthand part of theanswer

Consider:285×315(subtract 300 from eachnumber)

We will scale up thisintermediate result bymultiplying300by3.

Since 300 × 3 = 900,theanswerbecomes900/(–225).

Sincetheprimarybaseis 100, we will carry 3

from the left which isequal to a carry of 300(primary base 100 * 3)andsubtract225from300giving75.

So the final answerbecomes897/75=89775.

Conclusion

This section gives anidea of how a variety of

combinations of primaryand secondary bases canbe used to arrive at thefinal answer quickly.Thereader should choose thesecondary base carefullywhichcanreducethetimefor computationdrastically.

5. Examples formultiplication ofnumbers below the

base(100)

6. Examples for

multiplication ofnumbers above thebase(100)

7. Examples for

multiplication ofnumbers above andbelowthebase(100)

8. Examples for

squares of numbersbelowthebase(100)

9. Examples for

squares of numbersabovethebase(100)

10. Examples formultiplication ofnumbers(3-digits)below

thebase(1000)

11. Examples for

multiplication ofnumbers (3-digits) abovthebase(1000)

12. Examples for

multiplication ofnumbers(3-digits)above&belowthebase(1000)

13. Examples for

squares of numbers (3-digits) below the base(1000)

14. Examples for

squares of numbers (3-digits) above the base(1000)

15. Examples for

multiplication ofnumbers(4-digits)belowthebase(10000)

16. Examples for

multiplication ofnumbers(4-digits)abovethebase(10000)

17. Examples for

squares of numbers (4-digits) below the base(10000)

18. Examples for

squares of numbers (4-digits) above the base(10000)

19. Examples for

multiplication of mixednumbers

MULTIPLICATION(URDHVATIRYAK)

Vertically andCrosswise

Wewillnowconsiderthe other method ofmultiplication viz. the‘UrdhvaTiryak’or cross-multiplication technique.It consists of vertical andcross multiplicationbetweenthevariousdigitsand gives the final resultinasingleline.

We will consider themultiplication of thefollowing:

Two2-digitnumbersTwo3-digitnumbersTwo4-digitnumbers2-digit with 3-digitnumber2-digit with 4-digitnumber3-digit with 4-digitnumber

Once the technique isclear, it can be extendedto the multiplication ofanynumberofdigits.

Case 1 : 2-digitmultiplication

Consider thefollowing multiplicationasdoneby the traditionalmethod.

The Vedic mathsmethod consists ofcomputing three digits a,b and c as shown in thediagram

Steps:

Digit (a) : Multiplythe right most digits8and2toget16.

Digit (b) :Carryoutacrossmultiplication

between3and2,and8 and5 to get 6 and40 respectively andon adding them, weget46.

Digit (c) : Multiplythe leftmostdigits3and5toget15.

Wehavetoretainonedigit at each of the threelocations and carry the

surplustotheleft.

Hencein

15/46/16

we start from theright,retain6andcarry1to the left. This gives 47inthemiddlelocation.

Now,we retain 7 andcarry4totheleft.Weadd4to15toget19.

This gives 19 / 7 / 6i.e. 1976 which is therequiredanswer.

To speed up thisoperation, the carry digitcan bewritten just belowthe number on the left. Itcan then be added to thenumbertoit’sleftateachoperation itself. Let’s seethemultiplicationagain.

Here, the right mostvertical multiplication iscarriedoutfirsttoget8×2 = 16. We write 6 andwrite the carry 1 asshown.

Nextwecarryout thecrossmultiplication3×2

+ 5 × 8 = 46, add thecarry (1) to it andget47.Wewritedown7and thecarrydigit4asshown.

The3rdproductis3×5whichgives15towhichweaddthecarrydigitof4andget19.

As you can see, themethod gives you thefinal answer faster in just

oneline.

Let’s try anotherexample.

Steps:

Digit (a) : Multiplythe right most digits

6and3toget18.Wewrite 8 and show 1as the carry on it’sleft.

Digit (b) :Carryoutacrossmultiplicationbetween7and3,and6and4toget21and24 respectively andadd them to get 45.Addthecarrydigit1and get 46. Write

down6andshowthecarryas4.

Digit (c) : Multiplythe left-mostdigits7and4toget28.Addthecarrydigit4 to ittoget32.

20. Examples for 2digitcrossmultiplication

Case 2 : 3-digitmultiplication

Let us now considermultiplication of two 3digitsnumbers.

Consider thefollowingexample

The Vedic maths

method consists ofcomputingfivedigitsa,b,c, d, e as shown in thediagram

We will use themethod explained in thepreviouscaseinwhichthecarry digit at each stepwas added to the numbertoitsleft.

Steps:

Digit (a) : Multiplythe right most digits6and8toget48.Wewrite 8 and show 4as the carry, on itsleftside.

Digit (b) :Carryouta 2-digit crossmultiplicationbetween 2 and 8

(16), and 6 and 5(30).Add the products16 and 30 to get46.Addthecarrydigit4andget50.Write down 0 andshow the carry as5.

Digit (c) :Carryouta 3-digit cross

multiplicationbetween 3 and 8(24), and 6 and 2(12) and 2 and 5(10).Add the threeproducts 24, 12and10toget46.Addthecarrydigit5andget51.Write down 1 andshow the carry as

5.

Digit (d) :Carryouta 2-digit crossmultiplicationbetween3and5(15)and2and2(4).Add 15 and 4 toget19.Addthecarrydigit5toget24,Write down 4 andshow 2 as the

carry.

Digit (e) : Multiplythe leftmostdigits3and 2, to get 6.Addthecarrydigit2 to ittoget8.

Finally, themultiplicationlooksasfollows:

Let’s try anotherexample; the computationisasshown.

Case 3 :Multiplying3-digit and 2-digitnumbers

Ifwehavetomultiplya 3-digit with a 2-digitnumber, we can treat the2-digit number as a 3-digit number by paddingitwithazeroontheleft.

E.g. 354 × 45 can bewritten as 354 × 045 andthesametechniquecanbeused.

21. Examples for 3

digitcrossmultiplication

22.More examples 3

digits×2digits

Case 4 : 4-digitmultiplication

The techniquesdiscussed till now can beextended further for fourdigits.Thediagrambelowcanbeused

Let’s consider thefollowingexample

The final answer is9073792.

Steps:

Digit (a) : Multiplythe right most digits2 and 6 to get 12.Write down 2 andshowthecarryas1.

Digit (b) :Carryouta 2-digit crossmultiplicationbetween 3 and 6(18), and 2 and 5(10)andaddthemtoget28.Addthecarry

digit 1 and get 29.Write down 9 andshowthecarryas2.

Digit (c) :Carryouta 3-digit crossmultiplicationbetween1and6 (6),2and2(4)and3and5 (15) and add themto get 25. Add thecarry digit 2 and get27. Write down 7

and show the carryas2.

Digit (d) :Carryouta 4-digit crossmultiplicationbetween 2 and 6(12), 2 and 4 (8), 1and5(5)and3and2(6) and add them toget31.Addthecarrydigit 2 to get 33.Write down 3 and

show3asthecarry.

Digit (e) : Carry outa 3-digit crossmultiplicationbetween 2 and 5(10), 3 and 4 (12)and 1 and 2 (2) andadd them to get 24.Addthecarrydigit3and get 27. Writedown7andshowthecarryas2.

Digit (f) : Carry outa 2-digit crossmultiplicationbetween2and2 (4),and 1 and 4 (4) andadd them to get 8.Addthecarrydigit2and get 10. Writedown0andshowthecarryas1.

Digit (g) : Multiplythe leftmostdigits2

and 4 to get 8. Addthecarrydigit1 to ittoget9.

Let’s try anotherexample; the computationisasshown:

The final result is19745313.

Case 5 :Multiplying4-digit and 3-digitnumbers

Ifwehavetomultiplya 4-digit with a 3-digitnumber, we can treat the3-digit number as a 4-digit number by paddingitwithazeroontheleft.

E.g. 3542 × 453 canbewrittenas3542×0453

and the same techniquecanbeused.

23. Examples for 4digit and 5-digit crossmultiplication

24. Examples for

multiplication 4-digitsand5-digitsw3-digits

DIVISION

The operation ofdivision is basically the

reverse of multiplication.In the traditionalapproach, it isaslowandtedious method with aconsiderable amount ofguesswork. At each step,an approximate quotientdigit is guessed,which isthen multiplied by thedivisor and the remainderis computed. If theremainder is negative or

contains additionalmultiples of the divisor,the quotient digit isrecomputed. As thenumber of digits in thedivisor increase, thedivision becomesprogressivelyslower.

Vedic maths providesa very elegant methodwherein the division isreducedtoadivisionbya

single digit inmost casesor at most to two smalldigitslike12,13etc.inafew cases. At each step,thedivisionbya1-digitor2-digitdivisorisfollowedbyamultiplication,basedon the ‘Urdhav Tiryak’approach and finally asubtraction.

The reduction in bothtime and mental strain

using the Vedic mathstechniqueisastounding.

Consider the exampleof dividing 87108 by 84which isa2-digitdivisor.The traditional method isshownbelow:

At each stage, wehavetomentallycarryouta trial division and thenwrite down the quotientwhich gives either a

positive or zeroremainder. This is a slowandtediousprocedureandhence the operation ofdivision is detested. Asthe number of digitsincreaseinthedivisor,theprocess becomes moreand more slow and timeconsuming.

TheVedicmethodhasbeen devised beautifully

in which the divisionshave been reduced to aone-digit division. Thisspeeds up the processconsiderablyandmakesitanenjoyableexperience.

Case 1 : Division byanumberwith a flag ofonedigit(noremainder)

Let’s consider thesame example again and

carry out division by theVedicmethod.

Divide87108by84

We will write thedivisor as 84 where 4 iscalled the flag. Thedivision will be carriedout by a single digit viz.8.

Steps:

Place a colon onedigit from the right(= number of digitsofflag)

Divide 8 by 8 to getquotient as 1 and

remainderas0

Carry the remainder0 to the next digit 7and get 7 (calledgrossdividend,GD)

Multiplythequotient1bytheflag4toget4andsubtractitfrom7toget3(callednetdividend,ND)

Divide 3 by 8 to getthequotientas0andremainderas3

Carry the remainder3 to the next digit 1andget31(GD)

Multiplythequotient0bytheflag4toget0andsubtractitfrom31toget31(ND)

Divide31by8togetthequotientas3andremainderas7

Carry the remainder7 to the next digit 0andget70.

Multiplythequotient3bytheflag4toget12 and subtract itfrom70toget58

Divide58by8togetthequotientas7andremainderas2

At this point, theinteger part of thefinal quotient hasbeen obtained sincewe have reached thecolon. All thebalance digits wouldbethedecimalpart

Carry the remainder2 to the next digit 8andget28

Multiplythequotient7bytheflag4toget28 and subtract itfrom28toget0

Since the differenceisnow0,thedivisionis completed andweget the answer as

1037

Importantpoints:

Ateachstage,thenetdividend is obtainedby multiplying thequotient in theprevious column bytheflagdigit

Thenetdividendwasalwayspositive

Later, we will seewhat adjustmentshave to be made ifthe net dividendbecomes negative inanycolumn.

SecondExample

Divide43409by83

Steps:

Place a colon onedigit from the right(= number of digitsofflag)

Divide43by8togetquotient as 5 and

remainderas3

Carry the remainder3 to the next digit 4andget34

Multiplythequotient5bytheflag3toget15 and subtract itfrom34toget19

Divide19by8togetthequotientas2and

remainderas3

Carry the remainder3 to the next digit 0andget30

Multiplythequotient2bytheflag3toget6andsubtractitfrom30toget24

Divide24by8togetthequotientas3and

remainderas0

Carry the remainder0 to the next digit 9and get 09. At thispoint,theintegerpartof the final quotienthas been obtainedsince we havereached the colon.Allthebalancedigitswouldbethedecimalpart.

Multiplythequotient3bytheflag3toget9andsubtractitfrom9toget0

Since the differenceisnow0,thedivisionis completed andwegettheansweras523

Try a) 43412 / 83andb)43403/73

Case 2 : Division byanumberwith a flag ofone digit (withremainder)

Consider the divisionof26382by77

The quotient is 342and the remainder is 48,which appears under the

column just after thecolon.Thefinalanswerindecimalformis342.623.

Steps:

The steps asexplained above arecarried out till wereach the columnafterthecolonwhichshows the remainderas48. Ifwewant to

get the answer indecimal form, wehave to carry outdivision further byusing the sameprocedure.

Divide48by7togetquotient as 6 andremainderas6

Now, we attach azero to the dividend

and write theremainder to its left,giving the nextnumberas60.

Multiplythequotient6bytheflag7toget42 and subtract itfrom60toget18

Divide18by7togetquotient as 2 andremainderas4

Carry the remainder4 to the next digit 0(attached as before)andget40

Multiplythequotient2bytheflag7toget14 and subtract itfrom40toget26

Divide26by7togetthequotientas3andremainderas5

This process can becarried out upto anydesired level ofaccuracy.

Case 3 : Divisionwithadjustments

Consider the divisionof25382by77

Steps:

Place a colon onedigit from the right(= number of digitsofflag)

Divide25by7togetquotient as 3 andremainderas4

Carry the remainder4 to the next digit 3

andget43(GD)

Multiplythequotient3bytheflag7toget21 and subtract itfrom 43 to get 22(ND)

Divide22by7togetthequotientas3andremainderas1

Carry the remainder

1 to the next digit 8andget18(GD)

Multiplythequotient3bytheflag7toget21 and subtract itfrom 18 to get -3(ND)

Now, we have apeculiar situationwherein the ‘NetDividend’ has

become negative.The process cannotgo any further. Wewill now have tomakeanadjustment.

Wewill go one stepbackward andinsteadofwritingthequotient as 3 whiledividing22by7,wewillwritedownonly2asthequotientand

carry forward abigger remainder of8(ie22-7*2)

Carry the remainder8 to the next digit 8andget88(GD)

Multiplythequotient2bytheflag7toget14 and subtract itfrom 88 to get 74(ND) which is now

positive

Carryouttheprocessin the remainingcolumns, making anadjustmentwheneverthe net dividendbecomesnegative.

Thefinalresult is329with remainder 49 or indecimal form, it is329.6363.

Case 4 : Division byanumberwith a flag of2digits

Let us now considerdivision by a 3 digitdivisor where the flagwouldhavetwodigits.

Consider the divisionof2329989by514

Steps:

Place a colon twodigits from the rightsince we now havetwodigitsintheflag

Divide23by5togetquotient as 4 andremainderas3

Carry the remainder3 to the next digit 2andget32(GD)

Now, since there aretwodigitsintheflag,we will use twoquotient digits fromthe previous twocolumns to carry outacrossmultiplicationwiththetwodigitsofthe flag. We will

then subtract theresult from the grossdividend (GD) tocompute the netdividend(ND).

Since the quotient 4has only one digit,wepaditwithazeroon the left andmultiply thequotient04 by the flag 14 asfollows:

We will subtract 4from the GD 32 toget28(ND)

Divide28by5togetthequotientas5andremainderas3

Carry the remainder

3 to the next digit 9andget39(GD)

Now, the previoustwo quotient digitsare4and5whicharemultipliedbytheflag14asfollows:

We will subtract 21fromtheGD39toget18(ND).

Repeat the processfor the remainingcolumns.

Case 5 : Division byanumberwith a flag of3digits

Let us now consider

division by a 4-digitdivisor where the flagwouldhavethreedigits.

Consider the divisionof6851235by65245246

Steps:

Place a colon three

digits from the rightsince now there arethree digits in theflag

Divide 6 by 6 to getquotient as 1 andremainderas0

Carry the remainder0 to the next digit 8andget08(GD)

Now, since there arethree digits in theflag, we will usethree quotient digitsfrom the threeimmediatelyprevious columnsandcarryoutacrossmultiplication withthethreedigitsoftheflag. We will thensubtract the result

from the grossdividend (GD) tocompute the netdividend(ND).

Since the quotient 1has only one digit,we pad it with twozeroesontheleftandmultiply thequotient001 by the flag 524asfollows:

We will subtract 5fromtheGD8toget3(ND)

Divide 3 by 6 to getthequotientas0andremainderas3

Carry the remainder3 to the next digit 5

andget35(GD)

Now, the previoustwo quotient digitsare 0 and 1.Wepadthem with one zeroand multiply by theflag524asfollows:

We will subtract 2

from the GD 35 toget33(ND).

Divide33by6togetthequotientas5andremainderas3

Carry the remainder3 to the next digit 1andget31(GD)

Now, the previousthree quotient digits

are 1, 0 and 5. Wemultiply by the flag524asfollows:

We will subtract 29from the GD 31 toget2(ND).

Repeat the processfor the remaining

columns.

25. Examples fordivision-flagof1digit

26. Examples for

division-flagof2digits

27. Examples for

division-flagof3digits

SIMPLESQUARES

Inthischapterwewill

look at two simpletechniques. In the firstone,wewill learnhowtofind the square of anumber which ends in 5,e.g.35.Inthesecondone,we will see how to findthe product of twonumbers which have thesame starting digit andwhichhavethesumofthelastdigitas10e.g.33and

37.

Case 1 : Numbersendingin5

Considerthesquareof35.The technique is verysimpleandconsistsoftwosteps.

Steps:

Multiply the first

numberviz.3by thenumbernexttoiti.e.multiply 3 by 4 andwritedowntheresultas12.Next, multiply theseconddigiti.e.5byitself to obtain 25and attach it at theend of the resultobtained in thepreviousstep.

Theresultis1225.

Example-1

Compute652

Step1:6×7=42

Step2:5×5=25

And the final answeris4225.

Example-2

Let’scompute1252

Step1:12×13=156

Step2:5×5=25

And the final answeris15625.

The product of 12 ×13canbeobtainedbytheNikhilam method asexplainedbefore.

28. Examples forsquares of numbersendingin5

Case 2 : Twonumbers starting with

same digit and endingdigitsaddingupto10

Consider the example23×27.

Now,23×27=621

The product can beobtained by using thesame technique as forsquaring numbers endingwith5.

Steps:

Since both thenumbers start withthe same digit,multiplyit(i.e.2)byits next digit (i.e. 3)toget6.

Now multiply thelastdigitsofboththenumbersviz.3×7toget21.

The final result isobtained by theconcatenation of the twonumbers i.e. ‘6’ + ‘21’ =621.

Example-1

Compute84×86

The result will beobtainedas

Step 1 : 8 × 9 = 72

(first digit ‘8’ multipliedbyitsnextdigit‘9’)

Step 2 : 4 × 6 = 24(product of last digit ofeach number) giving thefinalresultas7224.

Example-2

Compute92×0.98

The result can beobtainedas

Step1:9×10=90

Step2:2×8=16

which gives theintermediate answer as9016.

Now, place thedecimal point after twodigitsfromtherighttogetthefinalansweras90.16.

Important

The same techniquecan also be used forbiggernumberslike318×312 or 1236 × 1234,where the last digits addup to 10 and theremaining digits are thesame.

29. Examples forproduct of numbershaving ending digitsaddingto10

SQUAREOFANYNUMBER

There are manyproblems in which we

have to compute thesquare of a number, i.e.multiply a number byitself. E.g. square of 382,2.3672etc.

Vedic maths providesa powerful method tocomputethesquareofanynumberofanylengthwitheaseandgettheanswerinone line. It uses theconcept of ‘dwandwa’ or

duplexwhichisexplainedinthischapter.

We will start withsome basic definitions,which are the buildingblocks for thesetechniques.

I. Definition -DwandwaorDuplex

Wewilldefinea term

called ‘dwandwa’ orduplexdenotedby‘D’.

a)Duplexofa singledigitisdefinedas

D(a)=a2

D(3)=32=9,D(6)=62=36etc.

b) Duplex of twodigitsisdefinedas

D(ab)=2*a*bD(37)=2*3*7=42D(41)=2*4*1=8D(20)=0

c) Duplex of 3 digitsisdefinedas

D(abc) = 2*a*c+b2

Thiscanbederivedbyusing D(a) and D(ab)definedabove.

We pick up the digitsat the two extreme endsi.e ‘a’ and ‘c’ andcomputeitsduplexas2*a*c.

We then moveinwards and pick up theremainingdigit‘b’.

Weadd theduplexofbi.e.b2totheresulttogetthe duplex of the 3 digits‘a’,‘b’and‘c’.

Thus D(346) =2*3*6+42

=36 + 16 =52D(130) = 2*1*0+32=9

D(107) = 2*1*7+02=14.

d)Duplexof 4digitsisdefinedas

D(abcd)=2*a*d+2*b*c=2(a*d+b*c)

Once again, we startfrom the two digits ‘a’

and ‘d’, compute theirduplex as 2 * a * d, thenmove inwards, we getanother pair ‘b’ and ‘c’and compute their duplexas2*b*c.

Thus the final duplexis2*a*d+2*b*c

D(2315)=(2*2*5)+(2*3*1)

= 20 + 6 =26

D(3102)=(2*3*2)+(2*1*0)

= 12 + 0 =12

D(5100)=(2*5*0)+(2*1*0)

=0

e) Duplex of five

digitsisdefinedasD(abcde) = 2ae+2bd+c2D(21354) = ( 2×2×4)+(2×1×5)+32

=16+10+9=35

D(31015) = ( 2×3×5)+(2×1×1)+02

= 30 + 2 +

0=32

Summary ofcomputationofduplex

30. Compute theduplex of the followingnumbers

II. Square of anynumber

Once the concept of‘duplex’ is clear, we cancomputethesquareofanynumberveryeasily.Letus

startwith the square of a2-digit number and thenextend it to biggernumbers.

Thesquareofa2-digitnumberabisdefinedas

(ab)2=D(a) /D(ab) /D(b)

Considerthesquareof34.

Then, 342 = D(3) /D(34)/D(4)

Where D(3) = 9,D(34)=24andD(4)=16

So,342=9/24/16.

Now, we start abackward pass from theright, retain one digit ateach location and carrythesurplustotheleft.The

working would look asfollows:

342=9/24/16= 9 / 25 / 6 ..Weretain6andcarry 1 to theleftgetting25.= 11 / 5 / 6 ..Weretain5andcarry 2 to theleftgetting11.

So the final answer is1156.

Similarly, 822 =64/32/4

=6724

The technique cannow be extended to a 3-digitnumberasfollows:

(abc)2 = D(a) /D(ab)/D(abc)/

D(bc)/D(c)Thus,2132=4/4/13

/6/9=

45369

The square of anynumber can now beobtained very easily byjust extending the set ofduplexes.

31. Examples for

squares of 2-digitnumbers

32.Squaresof3-digit

numbers

33. Examples for

squares of 4-digitnumbers

34. Examples forsquares of 5-digit

numbers

SQUAREROOTOFANUMBER

In the previouschapter we have seen theconcept of ‘duplex’ andhow it helps us tocomputethesquareofanynumberveryrapidly.

The computation ofthe square root alsoinvolves the use of theduplex,onlythistime,wewouldbe subtracting it ateach step to compute the

squareroot.

Wewillnowstudythetechnique to find thesquarerootofanynumberwhichmayormaynotbeaperfectsquare.Thefirstthree examples are ofnumberswhichareperfectsquares, and thenext twoexamplesareofimperfectsquares including adecimal number. The

general steps areexplained with specificexamplesbelow:

Steps:

Count thenumberofdigits,n,inthegivennumber.

If the number ofdigits is even, startwiththeleftmosttwo

digits, else startwiththe leftmost singledigit. Let us call thenumber so obtainedas ‘SN’ (startingnumber).

Consider the highestpossible perfectsquare,say‘HS’,lessthanorequalto‘SN’and compute itssquareroot‘S’.

Write down thesquareroot‘S’belowthe leftmost digits.This becomes thefirst digit of therequiredsquareroot.

Multiply the squareroot ‘S’ by 2 andwrite it on the leftside. This would bethedivisor,say‘d’.

For all the balancedigits in the originalnumber, the methodwould consist ofdivision by thedivisor ‘d’ aftersubtraction of theduplex of theprevious digits(exceptthefirstone).

The differencebetween ‘SN’ and

‘HS’ i.e. theremainder ‘R’ iscarried to the nextdigit and written toits lower left,forming a newnumber. This nextnumber is called thegrossdividend(GD).

The first GD isdivided by thedivisor ‘d’ to get a

quotient Q and aremainder ‘R’. Thequotient is writtendown as the seconddigit of the squareroot and theremainder is againcarried to the nextdigittoformthenextGD.

From this stageonwards, the duplex

of all digits startingfromtheseconddigitofthesquareroot,tothe column previousto the column ofcomputation, issubtracted fromeachGD to get the NetDividend,ND.

At each column, theND is then dividedbythedivisor‘d’and

the quotient iswritten down as thenext digit of thesquare root, and theremainder is carriedto the next digit toformthenextGD.

The process iscarried out till theND becomes zeroand no more digitsareleftinthenumber

for computation orwhen we havereached a desiredlevelofaccuracy.

If the ND does notbecomezero,wecanattach zeroes to theend of the givennumberandcarryouttheprocesstogetthesquare root to anylevelofaccuracy,i.e.

any number ofdecimalplaces.

Once the square rootis obtained or whenwedecidetostop,weplace the decimalpoint after therequired digits fromthe left. If n is even,the decimal point isput(n/2)digitsfromthe left. If n is odd,

the decimal point isplaced (n + 1) / 2digitsfromtheleft.

The process willbecome very clear onstudying the examplesgivenbelow.

I.Perfectsquareroot

Consider theexample of computing

the square root of 2116The working is asfollows which isexplainedstepwise.

Steps:

Count thenumberofdigits,n,whichis4.

Since n is even, westart with the twoleftmostdigitswhere‘SN’is21

Here the highestperfectsquare(‘HS’)lessthan21is16,itssquareroot(‘S’) is4and the difference‘R’is5.

We write the square

root 4 below theleftmost digits andwrite8(=2×4)onthe left side asshown.

4 is the first digit ofthe required squareroot.

For the balancedigits, the methodwould consist of

division by thedivisor8.

The difference i.e. 5is carried to thenextdigit1andwrittentoit’s lower left asshown. The value oftheGDisnow51.

TheveryfirstGDi.e.51 is divided by thedivisor 8 to get a

quotient 6 and aremainder 3. Thequotient is writtendown as part of thefinal result and theremainder is carriedto thenextdigit6 toformthenextGD36.

We have nowreached the thirdcolumn. We willsubtract the duplex

of the digit in thesecond column fromtheGDtogettheNetDividend,ND.

The digit in thesecondcolumnoftheresult is 6 and itsduplexis36.

36issubtractedfromtheGDwhichisalso36 to get the ND as

0. This signals theendoftheprocess.

Hence, the squareroot is obtained.Now, we will placethedecimalpoint.

Since n is 4 and iseven, the decimalpointisplaced(n/2)digits from the left.So,here,thedecimal

point will be placed2digits fromthe leftand the final answeris46.

II. Perfect squareroot

Consider theexample of computingthesquarerootof59049

Steps:

Count thenumberofdigits,n,whichis5.

Since n is odd, westart with theleftmostdigitviz.5

Here the highest

perfect square lessthan5is4,itssquareroot is 2 and theremainder‘R’is1(5–4).

We write the squareroot2belowtheleft-most digit and write4 ( = 2 × 2 ) on theleftsideasshown.

The remainder 1 is

carried to the nextdigit 9 giving thevalue of the GD as19.

19isdividedby4togetaquotient4andaremainder 3. Thequotient is writtendown and theremainder is carriedto thenextdigit0 togive the next GD as

30.

Theduplexof4is16which is subtractedfrom30 (GD) to get14(ND).

14isdividedby4togetaquotient3andaremainder 2. Thequotient is writtendown and theremainder is carried

to thenextdigit4 togivethenextGD24.

Now, the duplex ofthe two previousdigits4and3,i.e.2*4 * 3 = 24 issubtracted from theGD24togettheNDas 0. Since we havenot yet reached thelastdigitinthegivennumber, we cannot

stophere.On dividing 0 by 4we get the quotientas 0 and theremainder as 0,which is carried tothe next digit givingtheGDas09.

The threeimmediatelypreviousdigitsare4,3and0.Theirduplex

is9(2*4*0+32).When the duplex issubtracted, the NDbecomes equal tozero and no moredigits are left. Thissignalstheendoftheprocess.

Since n is 5, thedecimal point isplaced (n + 1 ) / 2digits i.e. 3 digits

from the left and thefinalansweris243.

III.Imperfectsquareroot as a decimalnumber

Consider theexample of computingthesquarerootof59064

Since this number isslightly larger than the

number taken in thepreviousexample,itisnota perfect square.Wewillnow see how to get itssquareroottoanydesiredlevel of accuracy.At anypoint, if theNDbecomesnegative,wegobackonestep and decrease thequotient by one toincrease the remainderthat is carried forward.

This process ofadjustmentisexplainedinthisexample.

Steps:

The first few stepsare the same till wereach the fourthcolumn where theGDis26andtheND

is2.

2 is divided by 4 togetaquotient0andaremainder 2. Thequotient is writtendown and theremainder is carriedto thenextdigit4 togivethenextGD24.

The duplex of 4, 3and 0 is 9 which is

subtracted from 24(GD)toget15(ND).

15isdividedby4togetaquotient3andaremainder 3. Weattach a zero tocontinue with theprocess.

The quotient iswrittendownandtheremainder is carried

to thenextdigit0 toformthenextGD30.

Now, the duplex ofthepreviousdigits4,3,0and3i.e.2*4*3 = 24 is subtractedfrom the GD 30 togettheNDas6.

On dividing 6 by 4we get the quotientas 1 and the

remainder as 2,which is carried tothe next digit givingtheGDas20.

The duplex of theimmediatelyprevious digits 4,3,0,3and1 is26(2*4 * 1 + 2 * 3 * 3)which gives anegativeND.Hence,we decrease the

previous quotient by1 and carry forwardthe remainder 6givingtheGDas60.

Now the duplex ofthepreviousdigits4,3,0,3and0is18(2* 3 * 3). When thisduplex is subtractedfromtheGD,wegettheNDas42.

This process can becarried out to anydesired level ofaccuracy.

Since n is 5, thedecimal point is put(5+1)/2=3digitsfrom the left, givingthe final answer as243.03086.

IV.Squarerootwith

adjustments

Consider the squareroot of a decimalnumber59064.78

This is a decimalnumber which is not aperfectsquare.

Steps:

Write the numberwithout the decimalpoint and carry outthe same steps asexplainedbefore.

The starting number(SN) is obtainedfrom the integerportonly of the decimalnumber.

The first few steps

are the same till wereach the fifthcolumn where theGDis24andtheNDis15.

As before, 15 isdividedby4togetaquotient 3 and aremainder 3. Wecarry the remainderto thenextdigit7 toget37(GD).

Now, the duplex ofthepreviousdigits4,3,0and3 is24(2*4 * 3) which issubtracted from theGD37togettheNDas13.

Ondividing13by4,we get the quotientas 2 and theremainder as 5,which is carried to

the next digit givingthe GD as 58. Notethat the quotient istaken as 2 and not 3to get a positiveNDinthenextcolumn.

The duplex of theimmediatelyprevious digits 4, 3,0,3and2 is34(2*4 * 2 + 2 * 3 * 3)which is subtracted

from 58 giving theNDas24.

24isdividedby4toget a quotient of 4andaremainderof8which is carriedforward. A zero isattachedat this stagetoenabletheprocesstocontinue tohigherlevelsofaccuracy.

This process iscarried further asshown.

Since n is 5, thedecimal point is put(5+1)/2=3digitsfrom the left, givingthe final answer as243.032467.

V. Perfect squareroot of a decimal

number

Now, consider thesquarerootofadecimalnumber 547.56 which isaperfectsquare.

Steps:

The stepsare similar

to those explainedbeforeandthereadershould have nodifficulty incomputingthesame.

Since the originalnumber has 3 digits,the decimal point isplaced after 2 digitsfromtheleft.

Important

If the starting numberSNissmall,sayless than4,thenthenextgroupof2digits can be included intheSN.e.g. incomputingthe square root of 13251,we can take SN as 132insteadof1.

35. Examples forsquare roots ofMiscellaneousnumbers

CUBESANDCUBEROOTS

a) Computing cubesof2-digitnumbers

Let us consider a twodigit number say ‘ab’.Further, let us considerthe expansion of theexpression(a+b)3,whichcan be used to find thecubeof thegivennumber‘ab’.

Weknowthat(a+b)3

=a3+3a2b+3ab2+b3

Wenotice that the1sttermisa3,

the2ndterma2b=a3×(b/a),the3rdtermab2

= a2b × (b/a)andthe 4th term b3

=ab2×(b/a)

Thus,eachofthe2nd,

3rd and 4th terms can beobtainedfromitspreviousterm bymultiplying it bythecommonratio(b/a).

We can also considerthe2ndterm3a2b=a2b+2a2b

andthe3rdtermas3ab2 = ab2 +2ab2

i.e. we can split it as

thesumoftwoterms.

Hence, ifwecomputea3 and the ratio (b/a),wecan derive all theremaining terms veryeasily. The two middleterms have to be thendoubled and added asshowntoget(ab)3.

Hence, inorder togetthe cube of any twodigitnumber, we start bycomputingthecubeofthefirst digit and the ratioofthe 2nd digit by the 1stdigit. The ratio can bemorethan,equaltoorlessthan one. The methodremains the same in allcases.

Case 1 : Ratio is

morethan1

Let us consider asimpleexamplei.e.123

Herethefirstdigitis1anditscubeisalso1.Theratioof the2nd to the1stdigitis2(=2/1).

Theremaining3termscan be obtained bymultiplying each of theprevious terms by 2 as

shown:

123=1248(cubeoflastdigit)

We also notice thatthe4thtermisthecubeofthe 2nd digit i.e. 8 is thecube of 2 which is thesecond digit in the givennumber12.

We have to now taketwicethevalueofeachof

the twomiddle termsandwrite them down. Onaddition,we get the cubeofthegivennumber12.

Similarly, let us lookat253

Case 2 : Ratio isequalto1

Case3:Ratio is lessthan1

36. Examples forcubesofnumbers

b) Cube roots of 2-digitnumbers

Let’s consider thecubes of numbers from 1

to9.13=123=833=2743=6453=12563=21673=34383=51293=729

Ifweobserveclosely,

we find that the last digitin every cube is uniqueand does not repeat foranynumber(1to9).Thisproperty can be used tofind the cube root of anytwo digit number veryeasily.

Let’s find the cuberootof438976.

We group it into sets

of 3 digits each, startingfrom the right. So, wehavetwogroupsviz.

438976

Steps:

We look at the firstgroupfromtheright,i.e.976andcompareits last digit i.e. 6with the cubes ofdigits from 1 to 9

where the cube alsoends with 6. Thenumber 216 endswith6.

Wetakeitscuberootwhich is 6. The lastdigit of the requiredcuberootis6.

Now, we considertheremaining3-digitgroup i.e. 438. We

locate the highestcube less than 438.The required cube is343 since the cubeaboveitis512whichisgreaterthan438.

Wetakeitscuberootwhich is 7. The firstdigit of the requiredcuberootis7.

The required cube

rootis76.

This techniqueprovides an easy andpowerful method tocompute 2-digit cuberoots. It canbeusedonlywhenthegivennumberisaperfectcube.

37. Examples forcube roots of 2 digitnumbers

c)Computing fourth

power of 2-digitnumbers

A similar approachcan be used to obtain thefourth power of a two-digitnumber,i.e.(ab)4

Subsequenttothefirstterm, each of theremaining terms is

obtained by multiplyingthe previous term by b/a.Let us consider examplesfrom each of the casesviz. when the ratio ismore than, equal to andlessthan1.

Case 1 : Ratio ismorethan1

Let us consider asimpleexamplei.e.124

Herethefirstdigitis1and its fourth power isalso 1. The ratio of the2ndtothe1stdigitis2.

Theremaining4termscan be obtained bymultiplying each of theprevious terms by 2 asshown:

Case 2 : Ratio isequalto1

Let us consider asimpleexamplei.e.114

Herethefirstdigitis1and its fourth power isalso 1. The ratio of the2ndtothe1stdigitis1.

Theremaining4termscan be obtained bymultiplying each of theprevious terms by 1 asshown:

Case3:Ratio is lessthan1

Let us consider asimpleexamplei.e.214

Herethefirstdigitis2anditsfourthpoweris16.Theratioofthe2ndtothe1stdigitis1/2.

Theremaining4termscan be obtained bymultiplying each of theprevious terms by 1/2 asshown:

38. Examples forfourth power ofnumbers

TRIGONOMETRY

It is that branch ofmaths in which we study

the relationships betweenthe sides and angles oftriangles. Most of thereaderswould be familiarwiththebasicconceptsoftrigonometry which aretaught in schools andcolleges. We have learntabout the definitions ofsine,cosineandtangentofagivenangleinatriangleand how to compute any

of these ratios, if any oftheotherratiosisgiven.

Let us refresh theseconcepts by using thefollowing right-angledtriangle:

The varioustrigonometric ratios aredefinedasfollows:

1.sinA=b/c

2.cosA=a/c3.tanA=b/a4.cosecA=1/sinA

=c/b5.secA=1/cosA=

c/a6.cotA=1/tanA=

a/b

Someof theproblemsin trigonometry are asfollows:

givenanyoneof thesix ratios for anyangle in a triangle,compute any of theremainingratiosgivenanyoneof theratios in a triangle,compute theremaining ratios fortwicethegivenanglee.g. given sin A,computetan2A

givenanyoneof theratios, compute theremaining ratios forhalf the given anglee.g. given sin A,computetanA/2

Standardtrigonometric formulaeexist to carry out variouscomputations of thisnature.

A.Triplet

In Vedic maths wehaveaconceptofatripletwhich isveryeffective insolving all these types ofproblems. In the abovetriangle, a triplet isdefinedas

a,b,c

which are themeasuresofthetwosides

‘a’ and ‘b’ followed bythehypotenuse(‘c’)attheend.

Therelationc2=a2+b2holdstrueinthiscase.

If any two values ofthe triplet are given, wecancomputethe3rdvalueand build the completetriplet.

The ratios are nowdefinedasfollows

sin A = 2ndvalue / lastvalue

cos A = 1stvalue / lastvalue

tan A = 2nd

value/1stvalue

Let us see someexamplesofhow tobuildandusethetriplet.

1. Suppose that a = 3andb=4.Letusseehowtobuildthetriplet.

Givenpartial triplet is3,4,

Therefore the lastvaluewillbe

Hence thecompletedtripletis3,4,5.

2. If we are given anincomplete triplet as 12,__,13,

The completed tripletwouldbe12,5,13wherethevalue5isobtainedas

Once the triplet isbuilt, all the sixtrigonometric ratios canbereadoffeasilywithoutany further computationoruseofanyformulae.

B. Computingtrigonometricratios

Letusassumethatina

triangle, the value of tanA is given as 4/3. Wehavetofindout thevalueof cosec A. Thetraditional trigonometricmethod would use thefollowingformula:

cosec2 A = 1 +cot2A

astanA=4/3,cotA=3/4

On substituting thisvalue in the givenformula,weget

cosec2 A = 1 +9/16

Therefore,cosec2A=25/16

andcosecA=5/4,sinA=4/5

If we also want the

value of cos A, we canusetheformula

tanA= sinA /cosA

On substituting thevalueof tanAandsinA,weget

cosA= sinA /tanA=(4/5)/(4/3)=3/5

Letusnowseehowtouse the Vedic mathstechniqueof the triplet tocompute the values ofcosecA,cosAetc.

The incomplete tripletinthisexampleis

3,4,__

tan A = 4/3 = 2nd

value/1stvalue

The complete tripletwouldnowbe

3,4,5

asseenbefore.

As soonas this tripletis built, we can read offalltheratios

E.g. cosec A = lastvalue/2ndvalue=5/4

cos A = 1stvalue / lastvalue=3/5

The reader can seethat there is no need tocarry out any complexcalculationsat all and thetriplet can be used toobtain all the six ratiosveryeasily.

C. Computing

trigonometric ratios oftwicetheangle

As we have seenabove, the triplet for theangleAiswrittenas

a,b,c

The triplet for theangle2Acanbeobtainedvery easily by using thefollowingcomputation

(a2 – b2), 2ab,c2

Letusseeanexample.

Assume that we aregiventhevalueofsinA=3 / 5 and we have tocomputetan2A.

The normal waywouldbeasfollows:

Sin2A+cos2A=1

Therefore cos2 A = 1–sin2A

=1–(9/25)

=16/25

ThereforecosA=4 /5

Now,sin2A=2* sinA*cosA

=2*(3/5)*(4

/5)

=24/25

Also, cos2 2A = 1 –sin22A

=1–(242/252)

=49/625

Thereforecos2A=7/25

andtan2A=sin2A/cos2A

=24/7

Consider the tripletmethodnow.

Since sin A = 3 / 5,thepartialtripletforangleAis__,3,5

And therefore thecomplete triplet for angle

Awillbe

4,3,5

The triplet for angle2A can be computed byusingtheformula

(a2–b2),2ab,c2

7,24,25

The value of tan 2Acan now be read off

directlyas

tan A = 24 / 7and

cos2A=7/24

Theeaseandeleganceof themethod is obvious.There is no need to useany formulae and all theratios can be read offeasily.

D. Computingtrigonometric ratios ofhalftheangle

If the triplet for theangleAis

a,b,c

then the triplet for theangleA/2canbeobtainedby using the followingcomputation

LetusseetheexamplewherethevalueofsinA=3 / 5 and we want tocomputetanA/2.

Now as pertrigonometricformula,

sinA=(2*tanA/2)/(1+tan2A/2)

If we solve thisequation,weget

tanA/2=1/3

Let us see the tripletmethod in action Thecomplete triplet for angleAis4,3,5

ThetripletforangleA/ 2 can be computed byusingtheformula

and hence the tripletforA/2is

Now,we can readoffthevalueas

tanA/2=1/3,

Exercise-39

*

AUXILLIARYFRACTIONS

We have seen elegant

techniques provided byVedic maths for divisionof any number by anydivisor. The methodsmaketheprocesseasyandfast and coupledwith thetechniques of mishrank(Chapter13), the increaseinspeedisamazing.

We will now seeadditional techniqueswhich can speed up the

division of fractionalnumberswhere:

the dividend is lessthanthedivisorandthe divisor is smalle.g. having 2 or 3digitsandthe last digit of thedivisor ends with 1,6, 7, 8 and9E.g. ofsuch divisors are 19,29, 27, 59, 41, 67,

121etc.

Other divisors endingin2,3,4,5and7canbeconvertedtosuchdivisorsby multiplying by asuitablenumber.

Wewill see examplesofdivisorsendingineachofthedigitsfrom1to9togetagoodinsightintotheentireprocess.

I) Divisors endingwith‘9’

Let us start with thecase where the divisorendswith9.

E.g.19,29,39,89etc.

GeneralSteps

Add 1 to thedenominator and use

thatasthedivisor

Remove the zerofrom the divisor andplaceadecimalpointin the numerator atthe appropriateposition

Carry out a step-by-step division byusing the newdividend

Every divisionshould return aquotient and aremainder andshould not be adecimaldivision

Every quotient digitwill be used withoutany change tocompute the nextnumber to be takenasthedividend.Inall

other cases, thequotient digit wouldbe altered by a pre-defined methodwhich will beexplained in eachcase.

Example1:

Let us now see thedetailsfor5/29

Steps

Add 1 to the divisortoget30

The modifieddivisionisnow5/30

Remove the zerofrom thedenominator byplacing a decimalpoint in the

numerator giving themodified division as0.5/3

Wewillnotwritethefinal answer as0.16666but

We will carry out astep-by-step divisionexplainedbelow

Steps for step-by-

stepdivision

Divide5by3,togetaquotient(q)=1andremainder(r)=2

Write down thequotient in theanswer line and theremainder justbelowittoitsleftasshown

Now,thenextnumberfor division is 21 whichondivisionby3givesq=7 and r = 0. The resultnowlooksas

Repeat the division ateach digit to get the finalanswer to any desiredlevel of accuracy !! Thenextfewdigitsareshown

below.

On dividing 7 by 3,we get q = 2 and r = 1.Theresultnowlooksas

On dividing 12 by 3,we get q = 4 and r = 0.Theresultnowlooksas

Thefinalresultupto8digitsofaccuracyis

and the result of 5/29=0.17241379.

Example2:

Similarly, 88/89 willbecomputedas

Example3:Similarly,41/119 will becomputedas

So, 41 / 119 =0.34453781

Notes:

For all numbers

ending with 3, wecan convert them tonumbersendingwith9 by multiplying by3.

E.g. if the divisor is13,wecanconvertitto39bymultiplyingby3.

Hence, 3/13 =9/395/23=15/69.

Similary, fornumbersendingwith7, we can multiplyby7togetanumberendingwith9.

E.g. if the divisor is17, we can convert it to119bymultiplyingby7.

Hence, 5/7 =35/49,4/17=28/1195/27=35/189.

40. Examples forauxiliary fractions-divisorendingwith9

II. Divisors endingwith‘1’

Now, let us considerthe fractions where thedivisors are numbersendingwith1.

E.g.21,31,51,71etc.

GeneralSteps

Subtract 1 from thenumerator

Subtract 1 from the

denominator and usethatasthedivisor

Remove the zerofrom the divisor andplaceadecimalpointin the numerator atthe appropriateposition

Carry out a step-by-step division byusingthenewdivisor

Every divisionshould return aquotient and aremainder andshould not be adecimaldivision

Subtract the quotientfrom9ateachsteptoform the nextdividend

Example1:

Let us now see thedetailsfor4/21

Steps

Subtract 1 from thenumeratortoget3

Subtract 1 from thedivisortoget20

The modifieddivisionnowis3/20

Remove the zerofrom thedenominator byplacing a decimalpoint in thenumerator

The modifieddivisionis0.3/2

Wewillnotwritethefinal answer as 0.15but

We will carry out astep-by-step divisionasexplainedbelow

Steps for step-by-stepdivision

Divide3by2,togetaquotient(q)=1andremainder(r)=1

Write down thequotient in the

answer line and theremainder justbelowittoit’sleftasshown

Subtract the quotient(1) from9 andwriteitdown

Now,thenextnumber

fordivision is18 andnot11.

On dividing 18 by 2,wegettheq=9andr=0.Theresultnowlooksas

Subtract the quotient(9) from9 andwriteitdown

Now,thenextnumberfordivisionis0andnot9.

Repeat thedivisionasexplainedbefore.

On dividing 0 by 2,we get q = 0 and r = 0.Theresultnowlooksas

Thefinalresultupto8digitsofaccuracyis

and the result of 4/21=0.19047619

Example2

Let us now see the

detailsfor7/111

Since the pattern isrepeating, we can writedowntheansweras7/111=0.063063063

Notes:

For all numbersending with 3, we

can convert them tonumbersendingwith1 by multiplying by7.

E.g. if the divisor is13,wecanconvertitto91bymultiplyingby7.

Hence,3/13=21/915/23=35/161

Similary, for

numbersendingwith7, we can multiplyby3togetanumberendingwith1.

E.g. if the divisor is17,wecanconvertitto51bymultiplyingby3.

Hence,5/7=15/214/17=12/515/27=15/81

Once themethods areclear for divisors endingwith9andwith1,wecanmove over to otherdivisors ending with 8, 7and6.

41. Examples forauxilliaryfractions

III) Divisors endingwith‘8’

Let us consider caseswhere the divisor endswith8.

E.g.18,28,48,88etc.

GeneralSteps

Add 2 to thedenominator and usethatasthedivisor

Remove the zero

from the divisor andplaceadecimalpointin the numerator atthe appropriateposition

Carry out a step-by-step division asexplainedbefore

Every divisionshould return aquotient and a

remainder andshouldnotbecarriedout as a decimaldivision

The quotient and theremainderistakenasthe next basedividend asexplainedabove

Sincethelastdigitofthedivisoris8which

hasadifferenceof1from 9, we multiplythequotientby1andadd to the basedividend at eachstage to compute thegrossdividend.

Example1:

Let us now see thedetailsfor5/28

Steps

Add 2 to the divisortoget30

The modifieddivisionisnow5/30

Remove the zerofrom thedenominator byplacing a decimalpoint in the

numerator giving themodified division as0.5/3

Wewillnotwritethefinal answer as0.16666but

We will carry out astep-by-step divisionasexplainedbelow

Steps for step-by-step

division

Divide5by3,togetaquotient(q)=1andremainder(r)=2

Write down thequotient in theanswer line and theremainder justbelowittoitsleftasshown

Now, the basedividendis21

Add thequotientdigit(1) to the initial dividendtoget22.

The next number fordivision is 22 which ondivisionby3gives q=7andr=1.Theresultnowlooksas

Now, the basedividendis17

Add thequotientdigit(7)tothebasedividendtoget24

The next number fordivision is 24 which ondivisionby3gives q=8andr=0.Theresultnow

looksas

Repeat the division toget the final answer asshownbelow.

Thefinalresultupto5digitsofaccuracyis

and the result of 5/28=0.17856.

42. Examples forauxilliary fractions-divisorsendingwith8

IV) Divisors endingwith‘7’

Let us consider caseswhere the divisor endswith7.

E.g.17,27,37,57etc.

GeneralSteps

Add 3 to thedenominator and usethatasthedivisor

Remove the zero

from the divisor andplaceadecimalpointin the numerator atthe appropriateposition

Carry out a step-by-step division asexplainedbefore

Every divisionshould return aquotient and a

remainder andshould not be adecimaldivision

The quotient and theremainderistakenasthe next basedividend asexplainedbefore

Sincethelastdigitofthedivisoris7whichhasadifferenceof2

from 9, we multiplythequotientby2andadd to the basedividend at eachstage to compute thenextgrossdividend.

Example1:

Let us now see thedetailsfor5/27

Steps

Add 3 to the divisortoget30

The modifieddivisionisnow5/30

Remove the zerofrom thedenominator byplacing a decimalpoint in thenumerator giving themodified division as

0.5/3

Wewillnotwritethefinal answer as0.16666but

We will carry out astep-by-step divisionexplainedbelow

Steps for step-by-stepdivision

Divide5by3,togetaquotient(q)=1andremainder(r)=2

Write down thequotient in theanswer line and theremainder justbelowittoitsleftasshown

Now, the initialbase

dividendis21

Multiplythequotientdigit(1)by2andaddto the base dividend21toget23

The next number fordivision is 23 whichon division by 3givesq=7andr=2.Theresultnowlooksas

Now, the basedividendis27

Multiplythequotientdigit (7) by 2 to get14 and add it to thebasedividend(27)toget41

The next number fordivision is 41 which

on division by 3givesq=13andr=2. The result nowlooksas

At this point, notethatthequotientdigithasbecomea2-digitnumberviz.13.Likebefore, it ismultipliedby2toget

26whichisaddedasshownbelow:

The remainder ‘2’ isadded to the tensdigitof13and26.

Now, we divide 59by 3 to get thequotient as 19 and

the remainder as 2.Theworkingappearsasshownbelow:

19ismultipliedby2to get 38, which isadded as shownbelow:

Now, we divide 77by 3 to get thequotient as 25 andthe remainder as 2.Theworkingappearsasshownbelow:

25ismultipliedby2to get 50, which isadded as shownbelow:

Ondividing 95 by 3we get the quotientas 31 and theremainder as 2. Theworking appears asshownbelow:

We will now see

howtobuildthefinalresult by retaining asingle digit at eachlocation.The answerconsists of thefollowingnumbers:

0. 1 7 (13) (19) (25)(31)

Start from the right,retain 1, carry 3 tothelefttoget

0.17(13)(19)(28)1

Retain 8 from 28,carry 2 to the left toget

0.17(13)(21)81

Repeat to get thefinalresultas

0.17(15)181

0.185 181

The final result with6 digits of accuracyis

5/27 =0.185181.

43. Examples forauxiliaryfractions

V) Numbers ending

with‘6’

Let us consider caseswhere the divisor endswith6.

E.g.26,36,56,86etc.

GeneralSteps

Add 4 to thedenominator and usethatasthedivisor

Remove the zerofrom the divisor andplaceadecimalpointin the numerator atthe appropriateposition

Carry out a step-by-step division asexplainedbefore

Every divisionshould return a

quotient and aremainder andshould not be adecimaldivision

The quotient and theremainderistakenasthe next basedividend asexplainedbefore

Sincethelastdigitofthedivisoris6which

hasadifferenceof3from9,ateachstage,we multiply thequotient by 3 andadd to the basedividend to computethe next grossdividend.

Example1:

Let us now see thedetailsfor6/76

Steps

Add 4 to the divisortoget80

The modifieddivisionisnow6/80

Remove the zerofrom thedenominator byplacing a decimalpoint in the

numerator giving themodified division as0.6/8

Wewillnotwritethefinalansweras0.075but

We will now carryout a step-by-stepdivisionasexplainedbelow

Steps for step-by-stepdivision

Divide6by8,togetaquotient(q)=0andremainder(r)=6

Write down thequotient in theanswer line and theremainder justbelowittoit’sleftasshown

Now, the initialbasedividendis60

Multiplythequotientdigit(0)by3andaddto the base dividend60toget60.Thisisaredundant step heresince the quotientdigitiszero,butitisshown for

uniformity.

The next number fordivision is 60 whichon division by 8givesq=7andr=4.Theresultnowlooksas

Now, the basedividendis47

Multiplythequotientdigit (7) by 3 to get21 and add it to thebasedividend(47)toget68

The next number fordivision is 68 whichon division by 8givesq=8andr=4.Theresultnowlooksas

Repeat the divisionto get the finalanswer as shownbelow.

The final result upto8 digits of accuracyis

44. Examples for

auxiliary fractionsdivisorsendingwith6

VI.OtherDivisors

In the sections givenabove, we have seendivisorsendingwith1,6,7,8and9.Whatdowedowith divisors endingwith2, 3, 4 and 5? We canconvert such divisors, bya suitable multiplication,to divisors for which wehave seen Vedictechniques. Use the tablegivenbelowforguidance.

We can see from thistablethat ifaproperdigitis used to multiply thenumerator and thedenominator, we canobtain a fraction wherein

we can apply one of theknown techniques andderivetheanswerquickly.Sometimes,wemay haveto carry out themultiplicationtwicetogetitinastandardform.

45. Examples forauxiliary fractionsmiscellaneousexercises

MISHRANK(VINCULUM)

The technique ofmishrankisverypowerfuland provides us with amethod to convert digitsin a number which aregreater than 5, to digitsless than 5. After theconversion, all arithmeticoperations are carried outusing the convertednumber, which make theoperationverysimpleand

fast.

It is not necessary toconvertallthedigitsbutajudicious conversion ofcertain digits (above 5)can decrease thecomputation effortconsiderably. The readerhas to be patient whilelearning this totally newand fascinating techniqueand gain expertise in

order to use this methodeffectively to greatadvantage.

We will begin byseeing the method toconvertanygivennumbertoitsmishrankequivalentand back to its originalvalue.

a) Conversion toMishrank

Mishrank digits arecomputed for those digitswhicharegreaterthan5.

Thestepsinvolvedare:

Subtract the givendigit from 10 andwrite it with a baraboveit

Add one to the digit

ontheleft

Examples

The mishrank digithasanegativevalueinthe

number.

Thus, in the firstexampleabove,

15 isequivalentto150-2i.e.148

Similarly, 2 3 isequivalent to2305 -20=2285while1 isequivalent to 10000 -1112=8888

b) Conversion fromMishrank

Only mishrank digits,which are marked with abar, are reconverted backtotheiroriginalvalues,asfollows:

Subtract 1 from thenon-mishrank digitto the immediate leftofthemishrankdigit

andwriteitdown.Subtract themishrank digit from10andwriteitdown.Repeat for allmishrankdigits.

Inexample4,wehaveaseriesofmishrankdigitsand we can use the sutra‘All from9and last from10’ to convert to theoriginalnumber.

Mishrank can be usedvery effectively inoperations like addition,subtraction,multiplication, division,findingsquaresandcubes

ofnumbers.Wewill lookat some examples to seeitstremendouspower.

c) Application inmultiple addition andsubtraction

Let us consider thefollowingproblem:

232 + 4151 –2889+1371

This could either besolvedbyfirstadding232and4151,thensubtracting2889 from the sum andthen adding 1371 to theresult.

Or,wecouldadd232,4151 and 1371 and thensubtract 2889 from thesum.

Instead, we could

speeduptheoperationbyconverting thesubtractiontoanadditionoperationasshownbelow:

Letuswrite–2889as+

Here, each digit iswrittenasanegativedigit.

Now,theproblemcanberewrittenas

232+4151+ +1371

orwrittenvertically,itappearsas

Since this is a singleoperation, it is faster andcauses less strain to themind.Themishrankdigitsin the result can be

convertedtonormaldigitstoobtain the final answeras2865.

d) Application insubtraction

Example 1 :Considerthesubtractionof6869from8988

The subtractionoperation would be slow

andmentallytiring,withacarryatalmosteverystep.The operation can besimplifiedandspeededupby using either of thefollowingtechniques:

Convert thesubtraction toadditionofanumberas shown in thepreviousexample.

Carry out a digit-by-digit subtraction,withoutanycarry.

Let’s see the firsttechnique.

6869 iswrittenwithabarontopofeachdigittosignifythatitisanegativedigit.

which is equal to2119. Here, in the unit’splace,8+ gives .

Let’s see the secondtechnique.

The same result canbe obtained if we simplysubtracteachdigitwithout

acarry.

Let’s see theoperation:

Here, in the unit’splace, 8 – 9 gives ,which gives 2119 onconversion frommishrank.

e) Application inmultiplication

Example 1 :Consider themultiplicationof69with48.

The operation can becarried out by the cross-multiplication techniqueexplained in Chapter 5.The computation can be

simplified further if weconvert the non-mishrankdigits to mishrank digits.We will see the methodbelow.

69=7

48=5

Now, we will carryout a cross multiplicationofthesetwonumbers.

Example 2 :Consider themultiplication of 882 by297

The correspondingmishrankdigitsare

8 8 2 = 9 2 (We

convert the 8 in the ten’splace and do not convertthe 8 in the hundred’splace)

297=30

Now, we will carryout a cross multiplicationofthesetwonumbers.

f) Application indivision

ExampleI:Considerthe division of 25382by77

77canbewrittenas8, wherewe convert 7 inunit’s place to itsmishrankdigit.

Divisioniscarriedoutby the same technique asexplained before. It willbe seen that there is noneed to make any

adjustment at anycolumnand the procedure canprogressveryfast.

Thefinalresult is329with remainder 49, or indecimal form it is329.6363.

g) Application insquares

In order to get the

squareofanynumber,weneed to compute duplexvalues of a lot ofnumbers. If some of thedigits are above 5, thecomputationoftheduplexbecomes tedious.Consider the square of897. The computation ofthe duplex is quite time-consumingsincethedigitsare big. We will see the

magic of the mishrankdigitsnow.

Both are validmishrank equivalentnumbers.

Wewillnowcomparethe computation of thesquare of the givennumber for all these

numbers.

Let’s look at the nextconversion,i.e.1 0

asbefore.

We can see that thecomputations becomeincreasinglysimpler.

h) Application in

cubes

Consider the cube of89where893=704969

The computationwithout using themishrankwouldappearasfollows:

The computation isvery tedious and

cumbersome.

Let’s convert tomishrank and seewhat istheresult.

This is easier andfaster.

In each of theoperations it can be seenthat the operation on thecorresponding mishranknumber is both easy andfast. The reader has topractice to convertnumberstotheirmishrankequivalents and backagain,tofullyavailofthepowerofthistechnique.

46. Examples in useofmishrank

SIMULTANEOUSEQUATIONS

Everyone is familiar

with simultaneousequations wherein wehave to solve twoequations in twounknownssayxandy.

The normal methodconsists of multiplyingeach equation by asuitable number so thatthe co-efficients of eitherx or y become same inboth theequations,which

canthenbesubtractedoutoftheequations,leavingasingle equation in oneunknown.

Vedic maths providesa simplemethod to solvethe equations withoutgoingthroughthislengthyprocess and gives thevalues of x and y in asingleline.Thiswouldbesimilarto‘Cramer’sRule’

but is easier to rememberanduse.

Wewillconsiderpairsof equations each withtwo unknowns say x andy.

Example 4x+7y=5…(1)

3x+9y=10…(2)

Thenormalmethodtosolve such simultaneousequations is to eliminateany one unknown bysuitable multiplication ofeach equation by aconstant and then solvingfor the other unknownvariable. In the aboveexample, multiplyequation (1) by 3 andequation (2) by 4 and

subtract.

Whichgivesy=5/3.

On substituting thisvalueofyinequation(1),weget

4x+7*5/3=5

Giving4x=-20/3∴x=-5/3

We will now see thetechniques provided byVedicmaths.

Equations in 2unknowns can be dividedinto 3 categories and wewillseethetechniquesforeachofthem.

Whentheratioofthecoefficients of either

xoryisthesameasthatoftheconstants

Whentheratioofthecoefficients of x andyareinterchanged

All other types notfalling in these twocategories

Category1

Consider the example:

3x+4y=10

5x+8y=20

In this example, theratioofthecoefficientsofyis4/8=1/2.Theratioofthe constants is 10/20 =1/2.

Since both are equal,

the value of x in such acase is equal to 0. If theco-efficients of xwere inthesameratio,valueofywouldbezero.

Thevalueofycanbeobtained from the firstequationbysubstitutingx=0,givingtheequationas

4y=10

ory=5/2

Category2

Now, let us considerthe method to solvesimultaneous equationswhere the coefficients ofx and y are interchanged.Let us consider theexamplegivenbelow.

45 x – 23 y =113…(1)

23x–45y=91

…(2)

Wewill add and thensubtract the givenequations to get asimplified set as shownbelow

68 x – 68 y =204…(3)

22x+22y=22…(4)

Dividingeq (3)by68andeq(4)by22,weget

x-y=3…(5)

x+y=1…(6)

This is a trivial set,whichgivesthevaluesof

x = 2 and y =-1.

Category3

Let us consider themethod to solve anygeneral simultaneousequation. Let us considertheexample solvedat thebeginningofthischapter.

4x+7y=5…(1)

3 x + 9 y = 10…(2)

Let us represent this

as a set of coefficients asfollows

abc

pqr

Wherea=4,b=7,c=5,p=3,q=9,r=10

We start with thecoefficientatthecentreintheupper rowviz ‘b’andget the value of ‘x’ by

usingtheformula

Notice the crossmultiplication which isused extensively here inthisformula.

b * r and c * q arecross-multiplications,from the centre to the

right.

Similarly,b*panda* q are crossmultiplications, from thecentretotheleft.

Togetthevalueofy

Here,again,c*panda * r are cross-

multiplications while thedenominator is the sameasbefore.

47. Exercises forpractice

OSCULATORS

The concept of‘Osculator’ is useful to

check thedivisibilityofagiven number by divisorsending with 9 or 1 or amultiplethereof.

E.g. Check thedivisibility of 52813 by59or

of32521by31or

of35213by

13or17.

An osculator is anumber defined for anynumber ending in 9 or 1and is obtained from thenumber by a simplemechanism described inthischapter.

The use of osculatorswouldbeseverelylimitedif only these two

categoriesofnumbersviz.ending with 9 or 1 wereconsidered. Interestingly,numbers ending with 3and 7 can also convertedtonumbersendingwith1or 9 by a suitablemultiplication. The sametechniquescanbeusedforallsuchnumberstoo.

Osculators arecategorizedintotwomain

types viz. positive andnegative depending onwhether the number endswith9orwith1.

PositiveOsculators

The osculator of anumber ending in 9 isconsideredpositive.

Itisobtainedby

Dropping9

Adding 1 to theremainingnumber

Theosculatorfor19is2,for29itis3andfor59,itis6.

If the given numberdoesnotendin9butin1or3or7,wecanmultiplyit by 9, 3 or 7respectively, to convert ittoanumberwhichendsin

9 and then compute theosculator by dropping the9.

Examples are givenbelowofhowtocomputethe osculators for variousnumbers.

Once the osculator ofa number has beencomputed, it is a simplematter to check the

divisibility of any givennumberbyit.

Example : Let uscheck the divisibility of228by19.

Steps

Compute theosculator of thedivisor (19)which is2.

Digit1

Start with thecomputation fromthe right hand digit(8)ofthedividend

Multiply it by theosculatortoget16(8×2)

Digit2

Addit to thedigit toits left to get 18 (16+2)

Multiply it by theosculator to get 36(18×2)

Subtract maximumpossiblemultiples ofthe divisor from it.Hence, 19 can besubtracted from it to

get17.

Digit3

Addit to thedigit toits left to get 19 (17+2)

Since 19 is divisibleby 19, the given numberisalsodivisibleby19.

Computation of

positive osculators fordifferentnumbers

48. Examples forchecking the divisibilityofthefollowingby19

49. Examples for

checking the divisibilityofthefollowingby29

50. Examples for

checking the divisibilityofthefollowingby13

51. Examples forchecking divisibility –

miscellaneousexamples

NegativeOsculators

Let us now considernegativeosculators,whichare defined for numbersendingwith1.

To get the negativeosculators of a numberendingin1

Drop1

The remainingnumber is the

osculator

Theosculatorfor11is1,for21itis2andfor61,itis6.

Example : Let us

work out the divisibilityof2793by21.

Steps

Start with theosculator of thedivisor (21)which is2

Mark all alternatedigits from thesecond last digit as

negativebyplacingabarontop

Digit1

Start with thecomputation fromthe right hand digit(3)ofthedividend

Multiply it by theosculator to get 6 (3

×2)

Digit2

Addit to thedigit toits left to get 3 (6 +9)Multiply it by theosculatortoget6(3×2)Subtract maximumpossiblemultiples ofthe divisor from it.

We cannot subtractanythinghere.

Digit3

Add it to theprevious digit to get1(7+6)Multiply it by theosculatortoget2(1×2)

Digit4

Add it to theprevious digit to get0(2+2)Multiply it by theosculatortoget0(0×2)

Since0isdivisibleby21, the given number isalsodivisibleby21.

52. Examples forchecking the divisibility

by21

53. Examples for

checking the divisibilityby31

54. Miscellaneous

examples

APPLICATIONSOFVEDICMATHS

We now come to themost interesting chapterof thisbook, inwhichwewill learn how to applythe techniques learnt sofar. After reading andsolving the problems inthis chapter, the readerwill realize the immensebenefit that is derived byusing the simple andelegantmethods ofVedic

mathsinsolvingavarietyofproblems.

I have divided thischapterinto3parts.Inthefirst part, I have listedtwenty questions selectedfrom various competitiveexams. The reader isencouraged to solve theseproblems independentlyand track the time takenfor solving. This part is

followed by the solutionsto the problems, usingtechniques from Vedicmaths, including areference to the relevantchapter.Thesolutionsarebrief but sufficient tofollowandthereaderwilleasily grasp theapplication of thetechniques learnt so far.The ease and the

efficiency would beappreciated. In the thirdpart,Ihaveincludedsomemore problems forpractice which the readershould solve and gainexpertiseinthesubject.

SampleProblems

1) Which is the greatest 6-digit number which is aperfectsquare?

2) If the length and thebreadth of a square isincreasedandreducedby5% respectively, what isthe percentage increaseordecreaseinthearea?

3)Compute896×896–204×204

4) If all the digits innumbers 1 to 24 arewritten down, would the

number so formed bedivisibleby9?

5)Whatshouldbethevalueof * in the followingnumber so that it isdivisibleby11?

8287*845*381

6) Which of the followingnumbers is divisible by

99? 135792, 114345,3572404,913464

7)Compute781×819.

8) A gardener planted103041 trees in such away that the number ofrows were as many astrees in a row. Find thenumberofrows.

9)Findthevalueof5112.

10)Packetsofchalkarekeptinabox in theformofacube. If the total numberof packets is 2197, findthe number of rows ofpacketsineachlayer.

11)Givena+b=10, ab=1. Compute the value ofa4+b4.

12) The population in avillageincreasesatarate

of 5% every year. If thepopulation in a givenyear is 80000, what willbe the population after 3years.

13)Whatisthelengthofthegreatestrodwhichcanfitinto a room of length =24mt,breadth=8mtandheight=6mt.

14) The difference on

increasing a number by8% and decreasing it by3% is 407. What is theoriginalnumber?

15) Rs. 500 grows to Rs.583.20 in 2 yearscompounded annually.What is the rate ofinterest?

16) A number 19107 isdividedinto2partsinthe

ratio 2 : 7. Find the twonumbers.

17) A sum of Rs. 21436 isdivided among 92 boys.Howmuchmoneywouldeachboyget?

18)Evaluate13992.

19) Evaluate 397 * 397 +397*104+104*104+397*104

20)Ifthesurfaceareaofthesideofacubeis225cm2.Whatisthevolumeofthecube?

Solutions usingVedicmaths

1)Whichisthegreatest6-digitnumberwhichisaperfectsquare?

Solution:A6-digit

numberwillhaveasquarerootofexactly3digits.

Thelargest3-digitnumberis999.Hence,thesquareof999istherequirednumber.

999×999=998001 (ReferChapter1)

2) If the length and thebreadth of a square is

increased and reducedby 5% respectively,what is the percentageincrease or decrease inthearea?

Solution:Letthelengthandbreadthbe‘l’and‘b’respectively.Thenewlengthwillbe1.05landthenewbreadthwillbe0.95b.Hence,thenewareawillbe1.05l×0.95

b.Bynikhilam,weget

1.05×0.95=100/-25=0.9975

Thenewareais0.9975lbandhencetheareadecreasesby(1.0000-0.9975)whichis0.25%.Or,thedecreasecanalsobereadoffdirectlyastherightsideoftheproductviz100/-25.

3) Compute 896 × 896 -204×204

Solution:Theproblembelongstothetypea2-b2wherethefactorsare(a-b)and(a+b).Hencethesolutionis

(896-204)(896+204)=692*1100

=761200(ReferChapter

3)

4) If all the digits innumbers 1 to 24 arewrittendown,wouldthenumber so formed bedivisibleby9?

Solution:Thenumberwouldbe

123456789101112131415161718192021222324.

Thesumofthefirst‘n’naturalnumbersisn(n+1)/2.

Hence,thesumwouldbe24*25/2=300andtheremainderis3.Thereforethenumberisnotdivisibleby9.(ReferChapter2)

5) What should be thevalue of* in the

following number sothat it isdivisibleby11?

8287*845*381

Solution:Forthefirstnumber,thesumsofthealternatesetsare22and(20+*).Ifthevalueofthe'*'is2,thedifference

willbezeroandthenumberwillbedivisibleby11.Hence,'*'=2.

Forthesecondnumber,thesumsare4and(8+'*').Ifthevalueofthe'*'is7,thedifferencewillbe11andthenumberwillbedivisibleby11.Hence,'*'=7.

6) Which of the following

numbers is divisible by99?

135792,114345,3572404,913464

Solution:Thenumbershouldbedivisiblebothby9and11.

Navaseshofeachofthenumbersis0,0,7,0

So,3rdnumberisrejected.

Ofthebalance,only2ndnumberisdivisibleby11.(Referchapter3).

Hence,114345isdivisibleby99.

7)Compute781×819.

Solution:Usethebaseas

800inNikhilamandwriteas

781–19819+19

Hence,theproductwillbe800/-381.

Multiplytheleftpartby8toconverttotheprimarybaseof100,toget6400/–381

So,borrow4fromlefttogetthefinalansweras6396/19.

8) A gardener planted103041 trees in such awaythat thenumberofrows were as many astrees inarow.Findthenumberofrows.

Solution:Computethesquarerootof103041

Hence,thenumberoftreesineachrowis321.

9)Findthevalueof5112.

Solution:Mentally,using500asthebase,

Weget5112=522/121.

Now,converttothebase

ofthousandbydividingtheleftpartoftheanswerby2.

So,thefinalansweris261121.

10) Packets of chalk arekept in a box in theform of a cube. If thetotalnumberofpacketsis2197,findthenumberof rows of packets in

eachlayer.

Solution:RefertoChapter10,takethecuberootof2197whichis13.Hence,eachrowcontains13packets.

11)Givena+b=10,ab=1.Computethevalueofa4+b4.

Solution:Sincea+b=

10andab=1,

(a+b)2

= a2+b2+2ab

a2+b2

= 100-2=98

(a2+b2)2

= a4+b4+2a2b2

a4+b4

= (a2+b2)2-2a2b2

= 982-2

=9604-2=9602.(UseNikhilamtoget982orally)

12) The population in avillage increases at arate of 5% every year.If the population in a

given year is 80000,what will be thepopulation after 3years.

Solution:Finalpopulation

=80000(1+5/100)3

=80000(21/20)3

= 10*213

Therefore,thepopulationwillbe92610.

13) What is the length ofthe greatest rod whichcan fit into a room oflength=24mt,breadth= 8 mt and height = 6mt.

Solution:LetthelengthofthegreatestrodbeL.

Then,L2 = 242+82+62

= 576+64+36=676

Therefore,L = 26mt.

UsetechniquesinChapters8and9toget

thesquareof24andthesquarerootof676.

14) The difference onincreasinganumberby8%anddecreasingitby3% is 407.What is theoriginalnumber?

Solution:LetthenumberbeN.

Now,(1.08N-0.97N)=

407

Hence,0.11N=407(407isdivisibleby11)

GivingN=3700

Therefore,theoriginalnumberis3700.

15) Rs. 500 grows to Rs.583.20 in 2 yearscompounded annually.

What is the rate ofinterest?

Solution:583.20=500(1+r)2

116.64=(1+r)2

ByNikhilam,weknowthat116.64=10.82

Hence,requiredrateofinterestis8%.

16) A number 19107 isdivided into 2 parts intheratio2 :7.Findthetwonumbers.

Solution:Thefirstnumberwouldbe(2/9)*19107.

Thisisadivisionby9,refertoChapter2,wegettheansweras4246.

Thesecondnumberis19107-4246,usemishranktogetthesecondas14861.

17) A sum of Rs 21436 isdivided among 92boys.How much moneywouldeachboyget?

Solution:Thisisadivisionproblem,refertoChapter6.

18)Evaluate13992.

Solution:Usemishranktoconvertthegivennumberto

140

Now,140 2=1/9/16/ /

/0/1

=1957201

19) Evaluate 397 * 397 +397*104+104*104+397*104

Solution:Theexpressionis

(397+104)2

= 5012

=502/001(byNikhilam)

= 251/001

20) If the surface area ofthesideofacubeis225cm2,whatisthevolumeofthecube?

Solution:Theareaofanyonesideis225.

Therefore,eachsideis15cm.(Sincethenumberendswith25,squarerootendswith5,andtogetthefirstdigit,1×2=2)

Thereforethevolumeis153whichis3375cm3.

Problems forPractice

1)Ashopkeeperknowsthat

13% oranges with himarebad.He sells 75%ofthe balance and has 261oranges left with him.How many oranges didhestartwith?

2)Whatshouldbethevalueof * in the followingnumbers so that they aredivisibleby9?

38*1

451*603

3)Theareaofasquarefieldis180625mt2.Ifthecostoffencingit isRs10permetre, how much woulditcosttofenceit?

4) Find the number nearestto 99547 which isdivisible by 687 :100166, 99615, 99579,98928

5) In a parade, soldiers arearranged in rows andcolumns in such a waythat thenumbersof rowsisthesameasthenumberof columns. If each rowconsists of 333 soldiers,how many soldiers arethereintheparade?

6) One side of a square isincreasedby15%andtheother is decreased by

11%. What is the %changeinthearea?

7)Findthesquarerootof

530.38090.7

8) If 1000 is invested inbank and earns 5%interest, compoundedyearly, what is theamount at the end of

threeyears?

9) The difference incompound interestcompounded half-yearlyand simple interest at10%onasumforayearis Rs 25. What is thesum?

10)Computethefollowing:1/35,3/14

11)‘A’sellsaradioto‘B’ata gain of 10% and ‘B’sellsitto‘C’atagainof5%. If ‘C’ pays Rs 924for it, what did it cost‘A’?

12)Findthenearestnumberto77993whichisexactlydivisibleby718.

13)‘A’received15%higherthan‘B’and‘C’received

15% higher than ‘A’. If‘A’received9462,whichof the following did ‘B’and‘C’receive?

(8218,10032or8042,10881or1419,10881or8228,10881)

14)Findthegreatestandtheleast 5 digit numberexactlydivisibleby654.

15) Anil walks 120 metresnorth, 70 metres west,175metres south and 22metres east. Find hisdistancefromhisstartingpoint.

Answers

1)1200oranges

2)6,8

3)Side=425mt,cost=Rs.17000

4)99615

5)110889soldiers

6)Increaseof2.35%

7) Square roots are 23.03,0.8366

8)Rs.1157.625

9)Rs.10000

10)0.0285714,0.2142857

11)CostisRs.800

12)78262

13) ‘B’ received 8228, ‘C’received10881

14)10464,99408

15)Anilis73metresaway.