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NUMBERS Fundae:

1. 1..the product of 2 numbers is 7168...and the HCF is 16..find the numbers...

answer 1)

numbers are (16,44 or (64,112)

approach is simple

devide 7168 with 16*16we get quotient as 28

which can be represented as 1*28 or 4*7

(2*14 cannot be considered as it will give 32 hcf)

FUNDA

to solve these kind of problems, just follow these 3 steps

1. find the value of product/(HCF^2)

2. find the possible pairs of factors of value obtained in step 1

3. multiply the HCF with the pair of prime factors obtained in step 2

2..a number on being divided by 5 & 7 succesively leaves the remainders 2 & 4

respectively..find the remainder when the same number is divided by 35...

***** Ques 2***** Ramu is given Rs 382 in one-rupee coins. He has been asked to

allocate them into a number of bags such that any amount required between Re

1 and Rs 382 can be given by just handling out a certain number of bagswithout opening any of them. What is the minimum number of bags possible?

*****

***** a) 15***** b) 13

***** c) 11

***** d) 17

Answer : (A)

Soln: whenever such a problem is given the key usually lies in powers of 2

or 3. So i tried with powers of 2 ( since we cannot open a bag )the various bags he can have are, 1, 2, 4, 8, 16, 32, 64, 128, 1, 2, 4, 8,

16, 32, 64 ( since 382 = 255 + 127 )

that is 15 bags, so the answer is A

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***** Ques 3

***** What is the remainder when (2222)^5555 + (5555)^2222 is divided by 7

?*****

***** a) 3

***** b) 5***** c) 6

***** d) 0

Answer: (D)

Soln: the rem of a+b when it is divided by x is same as the sums of rems

of

a,b wrt to x. that means,rem = ( (2222)^5555 mod 7 + (5555)^2222 mod 7 ) mod 7

2222^5555 mod 7 = 3 ^ 5555 mod 7 , is same as 5ll'ly 5555^2222 mod 7 = 4 ^ 2222 mod 7 , is same as 2

thus the rem is (5+2) mod 7 = 0

Quote:

Chandoo,Can you tell me how you got 3 ^ 5555 mod 7 as 5 and 4 ^ 2222 mod 7 as 2 ??

here it goes buddy,

3^1 mod 7 = 33^2 mod 7 = 2

3^3 mod 7 = 6

3^4 mod 7 = 43^5 mod 7 = 5

3^6 mod 7 = 1

3^7 mod 7 = 3 ( Again )

So, i divided 5555 with 6 and the and the rem is 5, so 3^5555 mod 7 is same as, 3^5 mod

7 and that is 5 ( look at the above list )

the same theory applies to 4^2222 mod 7

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FUNDA ON CUBES:

Hi junta,

here is some funda on cubes ,

We usually find questions invovling cubes in CAT. like, a big cube

painted red is cut into 64 smaller cubes and find the no of smaller cubes

with no sides painted etc. etc.

lets assume the cube is devided into n^3 small cubes.

then,

no. of small cubes with ONLY 3 sides painted : 8( all the corner cubes )

no. of small cubes with ONLY 2 sides painted :

A cube is painted on 2 sides means, it is on the edge of the bigger cube

,and we have 12 edges, each having n cubes. but since the corner cubes

are painted on 3 sides, we need to neglect them. so in effect, for each side

we will have (n-2) small cubes with only 2 sides painted.

thus, then number is, 12 * (n-2)

no of small cubes with ONLY 1 side painted :

for each face of the cube ( 6 faces ) we have (n-2)^2 small cubes with

only one side painted. and we have 6 faces in total.

so th number is, 6*(n-2)^2

no of small cubes with NO sides painted :

if we remove the top layer of small cubes from the big cube we will end

up a chunk of small cubes with no sides painted.

this number will be equal to, (n-2)^3.

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Originally Posted by guy_next_door83Two Trains A and B start from stations X and Y towards Y and X respectively. After

passing each other, they take 6 hours and 48 minutes and 3 hours and 20 minutes to reach

Y and X respectively. If train A is moving at 49 kms/hr, the speed of train B is...>

ok..now consider the pt. X where the 2 trains met..now XB = 49 * (24/5) = 235.2

and AX = x * (10/3) = 10x/3

hence XB/x = AX/49

or

235.2/x = 10x/(3*49)

solving... x = 58.8

For this problem there is a straight fomula,

S2 / S1 = sqrt( t1 / t2 )

where S2 and S1 are speed of the trains 2 and 1 and t2 and t1 are the times they take

to reach their destinations after meeting.

By formula, S2 = 49 sqrt ( 34x3 / 50 ) = 69.98Kmph.

Anil's approach is exactly correct (leave the minor arith error in fraction conversion 6 4/5

= 34/5 and not 24/5) I somehow feel that remembering the formula also helps

Originally Posted by tackledude

7. Take a 99 digit number. It is created by using first 54 natural numbers. The number

thus is 1234567891011.....5354. This number is now divided by 8. The remainder is :

(1) 1 (2) 2 (3) 4 (4) 8

can u explain the above stated problem?

Very simple problem,

Just divide the last 3 digits by 8 .... 354/8 ... Remainder = 2 i.e. Option (2)

Incase it was 16, take the last 4 digits and so on.

Quote:

Originally Posted by convolutedsignal

hey how does one solve problems like say ..

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(36472)^(123!) * (34767)^(76!)

last digit for any a^x will follow a cycle of 4. i.e. the last digit of a^5 is same as of a.

as 123! and 76! are divisible by 4

2^4 * 7^4 = 6*1 = 6!

Number System Page 15

Quote:

Originally Posted by Ninja

Can sumbody plz solve this?

What is the remainder when [(6!)^7!]^13333 is divided by 13?

The answer is 1.

6! =720 = 12*12*5

Thus we have [ 12*12*5] ^(7!*13333)

which can be broken as [(12)^(7!*13333)]*[(12)^(7!*13333)]*[(5)^^(7!*13333)]

Taking the first term i.e, 12^(7!*13333) when divided by 13 rem is 1.Similarly 5^(7!*13333) = 25^(5!*7*3*13333) this when divided by 13 gives rem as 1.

Thus Net remainder =1*1*1=1.

Question:What is the remainder when [(6!)^7!]^13333 is divided by 13?

Soln: 6!=720dividing 6! by 13 we get 5 as remainder.

So the remainder= (5)^(7!X13333)

=(5)^(2n) {Here n=an even number =1x3x4x5x6x7x13333} [6 makes n even]

=(5^2)^nNow,

we know {(5^2)^n}-1 is divisible by 13

Therefore,

(5^2)^n={(5^2)^n}-1+1 leaves a remanider of 1.

Take care

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Q: One more prob!!

If I say the digital sum of 1256 is 1+2+5+6 = 14 viz in

turn1+4=5.

Then what is the digital sum of 13^256?

Quote:

Originally Posted by Eccentric

say the number A= (10a + b) digital sum = a+bnow for A^2 digial sum would be same as (a+b)^2

:: This is only true if you do it recursively till you get a single digit number

for 13^1 digial sum is 4

for 13^2 it would be 4*4 ~ 16 ~ 7 (169 ~ 16 ~ 7)

for 13^3 would be 7*4 ~ 28 ~ 10 ~ 1 (2197 ~ 19 ~ 10 ~ 1)

for 13^4 would be 1*4 ~ 4for 13^5 would be 4*4 ~ 16 ~ 7

.. (A cycle starts here with perod of 3)

256 = 3X + 1

digital sum = 4

Good Show!!

you can also adopt the following approach:

as you said 'coz its 10a+b = a+b (as long as sum of digits is concerned)

so, 13 = 1+3 = 4

also, 256 = 2+5+6 = 13 = 1+3 = 4

so, it becomes 4^4 = again, 256 = 2+5+6 = 13 = 1+3 = 4 - answer!

Originally Posted by Eccentric

Could you explain how 13^6 mod 9 = 1 ???

I feel if you explain here olny (instead of PM it will help a lot others too)

hi all!

Euler theorem says

a^f(x) mod x=1 when

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a,x are co prime & f(x) is the no of numbers less than x and co prime to

x.

(for a prime no x , f(x)=x-1 )..

this gives famous

fermats theorem

a^(x-1) mod x =1

if x is prime & a,x are co prime

here in our problem..13,9 are co prime..but 9 is not prime..so we use euler theorem & not

fermat theorem..f(9)=6 becoz 1,2,4,5,7,8 are co prime& lt 9 but not 3,6..

so 13^f(9)mod9=1 i.e 13^6mod9=1

Binomial Theorem: pg: 20, 21

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number-system-21.html

Quote:

Originally Posted by akash76543

helloa..

What is the least number that should be multiplied to 100! to make it perfectly divisible by3^50

i found this question in ascent education's site...it seems to me that the answer they ve given s wrong..plz

give ur suggestions..

the link ishttp://www.ascenteducation.com/india.../arith2910.php

iam sure this might have been answered earlier..in case u dont find..this is the

solution..

highest power of n in x! can be found as [x/n]+[x/n^2]+...

where[] means greatest integer function.

here highest power of 3 available in 100! will be[100/3]+[100/9]+[100/27]+[100/81]

=33+11+3+1=48..

so for 100! to be divisible by 3^50 we need to multiply 100! by 3^2=9

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Quote:

Originally Posted by digitaltejas

5) What is the remainder when 2050 x 2071 x 2095 is divided by 23?

Solution:

(89*23 +3 ) * (90*23 +1) * (91*23+2) = let the multiples of 23 be called A type...

then (A+3)(A+1)(A+2) = A+6

now (A+6) / 23 = K + (6/23)

hence the remainder is 6.............

Originally Posted by digitaltejas

Thanks vinit for your answer, Dood, I dint understand why did u take the multiples of 23as A, cause all are different numbers and also where did the K come from? Please can

you elaborate.

hmm,hey man , that A is a group of all those numbers which are multiple of 23 .. its not just

one number ...i m just telling all those numbers which are multiple of 23 as A only 4 my

convinience ....

And that K is a constant value

let me make it very simple for u ..........

(89*23 +3 ) * (90*23 +1) * (91*23+2)

----------------------------------------

23When u find the product of first two termsof the numerator

the only term that is not multiple of 23 will come as 3 ..rest all will

be multiples of 23.Let the result of sum of all the multiples be A ....

Now we have the product of 1st two terms as (A +3)..

now our numerator is reduced to (A + 3) * (91*23 + 2)again by goin as we did b4 .....we gewt the final numerator as (B + 2*3)

here againb is a multiple of 23 ...

hence the only term thats left and not a multiple of 23 is 6 ....so remainder is 6 .............

Hope i was able to clearify .....

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LONG DIVISION

rule:divide the greater no by smaller one ,divide the divisor by the remainder

,divide the remainder by the next remainder and this goes on until, no remainder is

left

and we can find one inference out of this:-

any number which divides each of the two numbers also divides their sum,difference

of any multiples of that numbers.

a nice example was given in the bookhcf of 42 n 70=hcf of 28 and 42.......y 28??...becoz 28=70-42....

i hope now things r clear.....this means that even if u follow the long division method u

need not solve it till the end....once u see that the remainder and the previous divisor havegot some common terms...stop there and try to obtain the hcf of those two numbers

directly.....

******************************how to solve this

what is the coeff. of x^987 in the expansion of{x+x^2+x^3+x^4+.............x^1200}^234

a good question and the funda that works behind is equally good..

Originally Posted by ajatThats a good problem. I think its a part of Multinomial Theorem. I just know the approach.....Firstcondense the above series as a sum of GP i.e.

[x(1-x^1200)/(1-x)]^234= x^234(1-x)^(-234)(1-x^1200)Now there was some direct formula to find out the required coefficient in case of Binomial expansion for

Negative Index which i cant recall

Plz let us know! Thanks

GENFUNCTIONOLOGY: -

Concept of generating functions can be applied to solve lot many problems related to permutationsand combinations.

Lets understand first what generating function is

GEN. FUNCTION IS THE ALGEBRIC REPRESENTATION OF COMBINATORICS EVENTS.

Suppose the event is tossing a coin. It can be represented as (h + t) where h means getting a headand t means getting a tail. Tossing the coin twice will have a generating function(h + t)^2 = h^2 + 2ht + t^2 and the generating function generates all the combinatory

possibilities when these 2 coins are tossed. Here the exponent of each term represents the eventand the coefficient represents the number of ways in which the event can happen.In this case h^2 means getting head both times and that can happen in 1 way only. 2ht means

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getting a head and a tail and that can happen in 2 ways.

Another example here (This was also a question in CL MOCK-6)

There are two cubes one of which is having 5 faces red and 1 face blue the other is colored in redand blue such that when the two cubes are rolled then probability of getting same color on top faceof both the dice is . How many faces on the other cube are colored red?

I represent the one cube as 5r + b and lets say the other cube is represented by xr + yb and bynow u must know that this means x number of red faces and y number of blue faces.Here x+y = 6The product (5r+b)(xr+yb) = 5xr^2 + (5y + x) + yb^2 generates all the possibilities when u throw

the two dice. Since probability that both the dice show the same face is therefore out of 36 cases,18 cases will be favorable that means 5x + y (i.e. total cases in which both the dice show the sameface) = 18. so far we have two equations and two variables which after solving give x=3 and y=3.

So generating functions can be applied to so many combinatorial problems.Suppose the question is that in how many ways can u put 5 identical balls in 3 boxes.We know how to solve this problem but we may not know the genfunctinology way of solving it.Here is that way

Following generating function can represent balls in a box.

(x^0 + x^1 + x^2 +x^3 + x^4 + x^5)

And for 3 boxes (x^0 + x^1 + x^2 + x^3 + x^4 + x^5)^3 {exponent 3 because there are 3boxes each having same generating function) and coeff. of x^5 in the expansion of it will give total

number of ways in which these balls can be put into 3 boxes. Why?

Answer to this why is your task to determine. I am not explaining it here because thats too trivialto be explained.

Now how to find the coefficient? Well use an artificial trick that u may not be aware of but afterusing this 5478 times youll become used to it.

(x^0 + x^1 + x^2 + x^3 + x^4 + x^5 + x^6 + )^3 is our new generating function becausecoefficient of x^5 in it will have same value as the previous generating function had. Again why?Because the terms that we have added to the previous genfunction (x^6, x^7, .) has no effect on

coeff. of x^5 . If more whys arise then please think about that.

Now we are concerned about the coefficient of x and not the value of x. So assume that |x| < 1.that makes the generating function 1/(1 x)^3 or (1 x)^(-3).

This function can be expanded using binomial theorem as

{1 + (-3)(-x) + (-3)(-4)(-x)^2 / 2! + + (-3)(-4)(-5)(-6)(-7)(-x)^5 /5! + .

so coeff of x^5 is 7!/(2!*5!)

Behold we got the same answer as we would have had we applied the balls and barriers that is thenormal way of solving this sum. But the question is that when this problem can be solved using aneasier method then why to follow this procrustean approach. The answer is we wont follow thisapproach to solve that problem.

Rather well use the easier approach whenever asked that what is the coeff of x^5 in the expansionof ( 1 + x^1 + x^2 + x^3 + x^4 + x^5 )^3 . We know that answer will be same as number ofways of putting 5 identical balls in 3 boxes

I hope that explains another method of solving the question I asked.Please try that question and confirm the answer. There is much more about generating functionsbut time and especially relevance does not allow that to be posted.

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L(n) means last digit of n

So it will be safe to writeZ(100) = L(4^10)*Z(20)....4^10 because there are exactly 10 blocks of 10 consecutive

numbers, one 4 for each of the blocks.....

and that makes it a recursive sequence

so Z(100) = L(4^10)*Z(20)Z(20) = L(4^2)*Z(4)

Z(4) = 4 and L(4^2) = 6

so Z(20)= L(4*6) = 4

L(4^10) = 6so Z(100) = L(4^10)*Z(20) = 6*4 = 4........

Hope things are clear now......

But wait it is not generalized as yet......what if we are supposed to deal with......

finding last non-zero digit of 8293!.......[x] means integral value of x.....

Z(8293) = L(4^829)*[8293/5]*L(8291*8292*8293)

ok thats the homework....

solve it further and generalize if u can........

*************************************************

Pythgorean triplets at it's best...Given any number you can easily find out the basicpythagorean triplets. like i m talking

about 3,4,5 9,40,41

when 3,4,5 is multiplied by 3 it yields 9,12,15 but that's a derived pythagorean triplets.Steps to find the basic pythagorean triplets.

case 1: when 1st number is odd

hypoteneus = (sq(1st number)+1)/2

3rd number = hypoteneus - 1

e.g

1st number: 3

hypo: (3^2+1)/2 = 53rd number = 5-1=4

1st number: 13hypo: (13^2+1)/2 = 85

3rd number: 85-1=84

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Case 2: when 1st number is even

Hypo: sq(1st number/2)+1

3rd number: hypo-2

e.g:

1st number: 4hypo: sq(4/2)+1 = 53rd number: 5-2=3

1st number: 8hypo: sq(8/2)+1 = 17

3rd number: 17-2=15

The Pigeon Hole Principle asserts: If we must put N+1 or more pigeons in N holes, then

some pigeon hole must contain two or more pigeons. Although it might look nothingmuch than common sense, this principle is very useful in various types of problems. Itallows us to sometimes draw quite unexpected conclusions in situations, when it even

seems that we do not seem to have enough information.

The Pigeonhole Principle, also known as Dirichlet's Box (or Drawer) Principle, states

that, given two natural numbers n and m with n > m, if n items are put into m

pigeonholes, then at least one pigeonhole must contain more than one item. Another wayof stating this would be that m holes can hold at most m objects with one object to a hole;

adding another object will force you to reuse one of the holes (provided that m is finite;

otherwise, see below on infinite sets). More formally, the theorem states that there does

not exist an injective function on finite sets whose codomain is smaller than its domain.

Examples:

A bag contains beads of two colors: black and white. What is the smallest number of

beads which must be drawn from the bag, without looking, so that among these beads

there are two of the same color.Solution: We can draw 3 beads from the bag. If there are no more than one of each color

among these, then there would be no more than 2 beads altogether. This, obviously,

contradicts to the fact that we have chosen 3 beads.On the other hand, it is clear thatchoosing 2 beads is not enough, since they may happen to be of the same color. In this

problem the beads are the pigeons, and the colors play the role of pigeon holes.

An easy example of the pigeonhole principle involves the situation when there are five

people who want to play softball (n = 5 objects), but there are only four softball teams (m

= 4 holes). This would not be a problem except that each of the five refuses to play on ateam with any of the other four. To prove that there is no way for all five people to play

softball, the pigeonhole principle says that it is impossible to divide five people among

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four teams without putting two of the people on the same team. Since they refuse to play

on the same team, at most four of the people will be able to play.

Although the pigeonhole principle may seem to be a trivial observation, it can be used to

demonstrate possibly unexpected results. For example, there must be at least two people

in London with the same number of hairs on their heads. Demonstration: a typical head ofhair has around 150,000 hairs. It is reasonable to assume that no one has more than

1,000,000 hairs on their head (m = 1 million holes). There are more than 1,000,000

people in London (n is bigger than 1 million objects). If we assign a pigeonhole for eachnumber of hairs on a head, and assign people to the pigeonhole with their number of hairs

on it, there must be at least two people with the same number of hairs on their heads.

Another example is: Presume that in a box there are 10 black socks and 12 blue socks andyou need to get one pair of socks of the same colour. Supposing you can take socks out of

the box only once and only without looking, how many socks do you have to pull out

together? When asked point-blank, people may sometimes unthinkingly give answers

such as "thirteen", before realizing that the correct answer is obviously "three". To haveat least one pair of the same colour (m = 2 holes, one per colour), using one pigeonhole

per colour, you need only three socks (n = 3 objects).

If there are n persons (at least two) who can arbitrarily shake hands with one another,

there is always a pair of persons who shake the same number of hands. Here, the 'holes'correspond to number of hands shaken. Each person can shake hands with anywhere from

0 to n 1 other people. This creates n possible holes, but either the '0' or the 'n 1' hole

must be empty (if one person shakes hands with everybody, it's not possible to have

another person who shakes hands with nobody, and vice versa). This leaves n people tobe placed in at most n 1 non-empty holes, guaranteeing duplication.

**************************

EULERs & FERMATs THEOREM:

This one is for ignorant buddies who find remainder problems and theorems related to it

hard to understand.

I believe that algebraic number theory is one of the most intriguing topics ofmathematics. Numbers have enchanted mathematicians for centuries and still there is

much to be explored. Here is a set of theorems that can be used to solve almost all the

problems related with finding remainders. Before starting lets prove that number ofprimes are infinite. Suppose I have found first few primes as p(1) = 2; p(2) = 3 and so on

up to p(n). Product p(1)*p(2)**p(n) is divisible by all the primes we have found so far.

That means p(1)*p(2)**p(n) + 1 is not divisible by any of the known primes. Thatmeans either this number is a prime or it is a composite number having prime factors

different from that are known (Caveat: -We have no right to say that this number is prime

unless we prove it. A number of coaching institutes say this number is a prime but this is

not necessarily true). Hence if we have first n primes we can find at least one more prime

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and that proves infinitude of primes. Wasnt it a simple way of proving something of

quite an elusive nature? Well thats the beauty of mathematics.

Why does ISO insist upon standardizing the methods and approaches of running a

business? Because when you follow same process again and again, you become

proficient at using the process that brings the efficiency and quality in products andservices. So if we follow a prescribed set of rules, we will become proficient at solving

even difficult remainder problems.

It is necessary to know how a remainder is formally defined. Suppose we have two

positive numbers a, b where a>b. a = b*q + R is true for infinite sets of integers (q, R).

Lets say that R is the residue (because we have not defined remainder as yet). We define

remainder as value of R that is numerically less than the divisor b and either negative orpositive. Or r < b. A sample statement When a number is divided by 29 leaves a

remainder 5. should be interpreted as N = 29x + 5. I wont explain the trivial cases that

are given in almost all the materials, as when a number is divided by two different

divisors, remainders are same then remainder when LCM divides the number gives thesame remainder etc. Here are the theorems you must know.

First thing we should be clear about is that whatever operation (addition, subtraction or

multiplication) happens on the numbers, same operation happens on the remainders of the

numbers from a given divisor. Suppose 2 numbers x and y when divided by divisor aleave remainders r and R i.e. x = am + r and y = an + R.

(x + y) = a(m + n) + (r + R) hence remainder when (x + y) is divided by a is either (r + R)

or remainder when (r + R) is divided by a ( in case r + R is greater than a).

Similarly (x y)%a = (r R)%a and also (x*y)%a = (r*R)%a. (read % as remainder fromdivisor).

I presume everyone knows the application of binomial theorem in finding remainders sowont explain here.

RULE 1. Chinese remainder theorem: - Do not panic by looking at the name. It is justanother buddy that helps in solving remainder problems. I will explain it using a sample

problem.

Q. A number when divided by 7 leaves remainder 5 and when divided by 11 leaves

remainder 3, find a general solution. There may be easier methods available for solvingthese questions but motive here is to understand the approach that more or less remains

the same even when we move towards questions of higher level. From given statement

we can say that number N = 7x + 5 = 11y + 3.

Or x = (11y 2)/7

Lets say that y when divided by 7 leaves a remainder r i.e. y = 7a + r. 11 leaves either 4or 3 as remainder from 7 (at times negative remainder makes the calculations easier).

Since x is an integer, it gives remainder 0 when divided by 7 therefore 4r 2 is divisible

by 7 since r is the remainder from 7, youll have to try values of r up to 6. At r = 4, 4r 2

becomes divisible by 7. So we can say here that y = 7a + 4. Putting the value of y in

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N=11y + 3 = 11(7a + 4) + 3 = 77a + 47. Putting a = 0, 47 is the first number of this type

and 77a + 47 is the general series of such numbers.

I hope you can understand the process. Now lets say that in the same question another

condition is given that when divided by 13, number leaves a remainder 6.

That means N = 13b + 6 along with the previous conditions. We have already solve it fortwo conditions and found the combined form as 77a + 47. We just need to solve this with

the new form 13b + 6.

77a + 47 = 13b + 6Or b = (77a + 41)/13.

Again using the same process.

77%13 = -1 (Usefulness of negative remainders)

41%13 = 2 (Positives are equally useful)That clearly means that a%13 = 2

(-1)*(2) + (2) = 0{I have done the same operation on remainders as done on numbs}

Therefore a = 13c + 2; and putting this value in 77a + 47 = 77(13c + 2) + 47

= 1001c + 154 + 47 = 1001c + 201.Whatever other remainder conditions are given, we can solve in same fashion.

Now you know CHINESE REMAINDER THEOREM.

You need practice to master it. Do you require questions on this? I dont think so. You

can make as many as you wish to.

RULE 2: FERMATS LITTLE THEOREM

Using the properties of infinite arithmetic progressions, Fermat proved the theorem thatfor a prime number p that is co prime with another number a; when a to the power

(p-1) is divided by p, remainder is 1. Or a^(p-1) % = 1. Dont forget that this theorem is

defined only for prime divisors. E.g. 2^4%5 =1; 3^10%11 = 1. Or in more general forma^{n(p-1)}%p = 1. So if the question is 2^1000%11, remainder will be 1 because power

of 2 is a multiple of (11-1).

What if divisor is not a prime? Then we have eulars generalization of Fermats rule.

That generalization says that a^e(n)%n = 1. Where e(n) is eulars number for divisor n.

Also a is co prime with the divisor p. Eulars number e(n) for divisor n is defined as

number of natural numbers less than n and co prime with it. How to find e(n)? SupposeI want to find eulars number of 1001. Prime factors of 1001 are 7,11,13.

e(1001) = 1001(1-1/7)(1-1/11)(1-1/13). In general for a number n having prime factors

p1, p2, p3. e(n) = n (1-1/p1)(1-1/p2)(1-1/p3)This knowledge about eulars theorem is sufficient. Now that we know fermats rule and

eulars rule, lets try a sample question.

What are last two digits of 3^96? Last two digits of this number can be obtained by

finding remainder when this number is divided by 100. 100 = (2^2)*(5^2). Understand

the things clearly here. For 2^2 i.e. 4, eulars number is 4(1-1/2) = 2 and for 5^2 (25),

eulars number is 25(1-1/5) = 20. Using eulars theorem, 3^2 %2^2 = 1 and

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3^20 %5^2 = 1. if we take LCM of the powers of 3 in two cases(LCM is 20) we find that

3^20 %100 = 1 or last two digits of 3^20 are 01. It can be proved for any other number co

prime with hundred. That explains why cyclicity of last two places is generally 20 for allnumbers co prime with 100. Last 2 places of 3^96 will therefore be same as last two

places of 3^16. (Because 3^96 = 3^80 * 3^16 and 3^80 ends in 01) . How to find last two

digits of 3^16? 3^20 = 3^16 * 3^4 lets say last two digits of 3^16 are xy.Therefore xy * 81 = 01 (considering only last two digits) thats true for xy = 21.

Another question. What is the remainder when 3^1001 is divided by 1001?

1001 = 7*11*13. 3^6 %7 = 1; 3^10 %11 = 1 and 3^12 %13 = 1. That implies

3^{LCM (6,10,12)} %1001 = 1 or 3^60 %1001 = 1. Using this we can reduce the

previous problem into a simpler one. 3^1000 = 3^960 * 3^41, from 3^960, remainder is 1therefore reminder when 3^1001 is divided by 1001 is same as remainder when 3^41 is

divided by 1001. 3^41 %7 = 3^5 %7 = 5 {I presume it should be clear by now}

3^41 %11 = 3

3^40 %13 = 3^5 %13 = 9. It is a number which when divided by 7,11 and 13 givesremainders 5,3 and 11 respectively. N = 7x + 5 = 11y + 3 = 13z + 9. Further we know

how to solve it (HINDI CHEENI BHAI BHAI).

I am attaching a list of questions few of which are collected from various threads and a

few are made by me. Lets solve all those using the standard methods we have learnt.Solving the problems will make concepts clearer. Remember; try to standardize your

approach.

RegardsSumit.

1.What is the remainder when 17^19 + 13^19 is divided by 25?2.Find the remainder when 104^303 is divided by 101.

3.Find the last two digits in the expansion of 2^ 999.

4.Find the remainder when 2 ^ 1990 is divided by 1990.

5.Remainder when (128 )^500 is divided by 153.6.Find last 3 digits of 3^1994.

7.What is the remainder when 2^2001 is divided by 2001?

8.What is the remainder when 2^2002 is divided by 2002?9.What is the remainder when 13^2404 is divided by 2310?

10.What is the remainder when 2^10013 is divided by 3125?

11. Find last 4 digits of (2319)^{10^12 + 2}.And a lengthy one here.

12. What is the remainder when 17^28820 is divided by 30030?

Solve these first and Ill post more related with the topic.

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