For KVS TEACHERS - WordPress.com · The required number is H.C.F of 384, 528, 144. The H.C.F is 48....

MATHS OLYMPIAD WORKSHOP

For

KVS TEACHERS

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Ans1. There are two possibilities- (i) If digits can be repeated- Largest number of six digits = 999999 Smallest number of six digits=100000 Difference=899999 (ii) If digits are not repeated- Largest number of six digits = 987654 Smallest number of six digits=102345 Difference=885305

Q1 Find the difference between the largest and the smallest numbers that can be formed with six digits.

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q2.The average of nine consecutive natural numbers is 81. Find the largest of these numbers.

Ans. Let numbers be x. x+1, x+2,x+3,x+4, x+5,x+6,x+7,x+8 Average=81 =>(9x+36)/9=81 => x+4=81 =>x=77 Hence Largest number =85

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q3.What will be 77% of a number whose 55% is 240? ANS. 55% of x=240 => x=240*100/55 77%of x=(77*240*100)/55*100=336

Q4.Flowers are dropped in a basket which become double after every minute. The basket became full in 10 minutes. After how many minutes the basket was half full?

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Ans. Flowers in basket become double after every minute. Therefore, Basket was half full 1 minute before it becomes full i.e. in 9 minutes.

Q5. A number consists of 3 digits whose sum is 7. The digit at the units place is twice the digit at the ten’s place. If 297 is added to the number, the digits of the number are reversed. Find the number.

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs )

Ans. Let, Digit at Hundred’s place=x Digit at Ten’s place=y. Then Digit at one’s place=2y. Also sum of digits=7 => x+y+2y=7 5 => x=7-3y Number=100(7-3y)+10y+2y=700-288y On reversing digits new Number=100(2y)+10y+(7-3y)=207y+7 Since on adding 297 digits are reversed Therefore, 700-288y+297=207y+7 => 495y=990 ,y=2 Hence, Number=124

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q6.When an integer ‘n’ is divided by 1995.The remainder is 75. What is the remainder when ‘n’ is divided by 57?

Ans.

Let, The Number be n, when divided by

1995 leaves remainder 75.

=> n=1995xq+75

=> n=57x35q+57+18

=> n=57(35q+1) +18.

Hence, remainder will be 18 when the

number is divided by 57. 6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q7.Find the missing digits in the following multiplication. 3 5 9 7

* * * ------------------------------ * * * * * * * * * * * * * * *** -------------------------------- * * * * 5 4 1 ---------------------------------

Ans. 3 5 9 7 x 7 5 3 ------------------------------ 1 0 7 9 1 1 7 9 8 5 2 5 1 7 9 -------------------------------- 2 7 0 8 5 4 1 ---------------------------------

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs )

Q8.Consider the following multiplication in

decimal notations (999)X(abc) = def132,

determine the digits a,b,c,d,e,f.

Ans. Given, 999 X abc = def132

LHS = (1000 – 1) abc

= abc000-abc

10 – c =2 ⇒c = 8

9 – b = 3 ⇒ b = 6

9 – a = 1 ⇒ a = 8

c – 1 = f ⇒ f = 7

d=a=8 & e=b=6

Q9.Find the largest prime factor of 314+313-12

Ans :- 314+313-12 =313 (3+1) -12 =3.4(312-1) =3.4(36-1)(36+1) =3.4.(32-1)(34+32+1)(32+1)(34-32+1) =3.4.8.91.10.73 =26.3.5.7.13.73 Largest prime factor of 314+313-12 =73

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q10.Find the greatest number of four digits which when divided by 2,3,4,5,6,7 leaves a remainder 1 in each case. Ans)

Required number will be 1 more than greatest

four digit multiple of 2,3,4,5,6,7 .

LCM of 2,3,4,5,6,7 =420

10000=420x23+340

Greatest four digit multiple of 2,3,4,5,6,7

=420x23=9660

Required number=9660+1=9661

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs

Q11.

Find the greatest number of four digits

which when increased by 1 is exactly

divisible by 2,3 ,4,5,6 and 7 Ans.

LCM of 2,3,4,5,6,and 7=420

Largest four digit number which is a multiple

of 420= 9660

Therefore required number = 9660-1

=9659

(Since required number is 1 less than the exact multiple)

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs )

Q12.Find the greatest number of 4 digits,

which when divided by 3,5,7 and 9 leaves

remainder 1,3,5 and 7 respectively.

Ans.

Greatest four digit number = 9999

LCM of 3,5,7 and 9 = 315

Highest four digit multiple of 315 = 9765

Required Number = 9765 – 2

= 9763

As 3 - 1 = 5 - 3 = 7 – 5 = 9 – 7 = 2

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs

Aliter:

Let, The Number = 3x + 1 = 5y + 3

= 7z + 5 = 9u + 7

= 3(x + 1) – 2 = 5(y+1) – 2=7(z+1) – 2=9(u +1)– 2

i.e. Number is 2 less than common multiple of

3,5,7 and 9.

L.C.M. of 3,5,7 and 9 = 315

Greatest no. of 4 digits = 9999 = 315×31+ 234.

Greatest number of 4 digits which is a

multiple of 315 = 9999-234=9765

Therefore, required number = 9765-2= 9763

Q13. Find the greatest number of five digits which is divisible by 56, 72, 84 and 96 leaves remainders 48, 64, 76 and 88 respectively. Ans. Let number be x X=56a+48=72b+64=84c+72=96d+88 =56(a+1)-8=72(b+1)-8=84(c+1)-8=96(d+1)-8 Number must be 8 less than a multiple of 56, 72, 84, 96 L.C.M of 56,72,84 and 96 = 2016 Greatest number of 5 digits =99999 =2016X49+1215 Largest multiple of 5 digits = 99999-1215= 98784 Required number = 98784 -8= 98776

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q14. A number when divided by 7,11 and 13(the prime

factors of 1001) successively leave the remainders

6,10 and 12 respectively. Find the remainder if the

number is divided by 1001.

Ans.Let, X= 7q1+6 = 7(q1+1)-1

X= 11q2+10 = 11(q2+1)-1

X= 13q3+12 = 13(q3+1)-1

Hence number is 1 less than common

multiple of 7,11 and 13

LCM of 7,11 and 13=1001

Hence, X=1001q-1 =1001(q-1)+1000

Therefore when X is divided by 1001 will

leave remainder 1000.

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q15. A number ‘X’ leaves the same remainder

while dividing 5814, 5430, 5958. What is

the largest possible value of ‘X’.

Ans. According to the given condition,

5814 = aX + r, 5430 = bX + r, 5958 = cX +r

This implies the difference of any of the above 3

numbers is divisible by X.

5814 – 5430 = 384, 5958 – 5430 = 528,

5958 – 5814 = 144.

The required number is H.C.F of 384, 528, 144.

The H.C.F is 48.

The required number here is 48.

Q16.How many prime numbers between 10

and 99 remain prime when the order of

their digits is reversed?

Ans.

There are 9 numbers between 10 and

99 which remain prime when the order

of their digits is reversed.

These are- 11,13,17,31,37,71,73, 79

and 97.

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q17. Exactly one of the numbers 234, 2345,

23456, 234567, 2345678, 23456789 is a

prime. Which one must it be?

Ans.

234,23456,2345678 are even.

2345 is divisible by 5

234567 is divisible by 3.

Hence, 23456789 is prime.

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q18. A two-digit number is such that if a decimal point is placed between its two digits, the resulting number is one- quarter of the sum of two digits. What is the original number?

Ans. Let, Number=10x+y If decimal is placed- (x+y)/10=1/4(x+y) y=5x Only possible value for x is 1 Therfore, Number=15

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q19. Which is greater: 31 11 or 17 14

Ans. 3111-1714=1711((31/17)11-173 ) =1711((31/17)11-4913 ) But 1<31/17 <2 =>(31/17)11<211

=>(31/17)11<2048

=>(31/17)11<4913

=>(31/17)11 - 4913< 0 3111-1714 < 0 1714 Greater Than 3111

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Without actually calculating, find which

is greater: 3111 or 1714.

Ans.

3111 < 3211 => 3111<(25)11

=> 3111<255

AND 1614 < 1714 => (24)14 < 1714

=> 256 < 1714

Hence 3111<255<256<1714

=> 3111<1714

Q20. Show that 199+299+399 +499+599 is exactly divisible by 5

Ans. 199+299+399 +499+599 =(199+499)+(299+399)+599

each is divisible by 5 {Since x n + y n is divisible by x + y when n is odd} 199+299+399 +499+599 is exactly divisible by 5

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q21. Find the number of perfect cubes between 1 and 1000001 which are exactly divisible by 7

Ans. Number of perfect cubes between 1 and 1000001, which are exactly divisible by 7 must be cubes of numbers between 1 and 100 that are exactly divisible by 7. Therefore, required number of such cubes = 14

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs

Q22. How many numbers from 1 to 50 are divisible by neither 5 nor 7, and have neither 5 nor 7 as a digit.

Ans.

Number of numbers divisible neither by 5 nor by 7

= 50 -10 – 7 +1=34

Numbers having 5 or/and 7 as digit in above numbers are 17 , 27, 37 and 47

Hence, Required number of numbers = 34 - 4=30

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q23.The square of a number of two digits is four times the number obtained by reversing its digits . Find the number. Ans. Let Number be 10x+y (10x+y)2 = 4.(10y+x) Number is even and 10y+x, IS A PERFECT SQUARE .

Possible values= 25,36,49,64 and 81 Square root of 4(10y+x) may be 10,12,14,16 ,18. 10 ,12,14 and 16 does not satisfy other conditions. Required number is 18.

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs

Q24.Find the sum of the digites in

2 2000 . 52004

Ans.

22000 .52004

= 54.22000.52000

= 625. 102000

Therefore , Sum of digits =6+2+5=13

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs

Q25.Find the last two (ten’s and unit’s) digit

of (2003)2003 Ans.

Last two digits is remainder when number is divided by 100

(2003)2 ≡ 32 (mod 100 ) ≡ 9 ( mod 100 ) (2003)4 ≡ 92 (mod 100 ) ≡ -19 (mod 100 )

(2003)8 ≡ (-19)2 (mod 100 ) ≡ 61 (mod 100 )

(2003)16 ≡ 612 (mod 100 ) ≡ 21 (mod 100 )

(2003)32 ≡ 212 (mod 100 ) ≡ 41 (mod 100 )

(2003)40 ≡ (2003)32 .(2003)8 ≡ 41.61 (mod 100 )

≡1(mod100)

(2003)2000 ≡ (200340)50 ≡ 150 (mod100) ≡ 1(mod100)

(2003)2003 ≡ 20032000.20032.20031(mod100)

≡1.9.3(mod100) ≡27(mod100)

Last two digits of 20032003 =27

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q26. Find the remainder when 22005 is divided by

13.

Ans.

22005=22000.25 , 25≡6(mod 13)

210= (25)2 ≡62(mod13) ≡10(mod13)

220= (210)2 ≡102(mod13)≡ 9(mod13)

240= (220)2 ≡92(mod13) ≡3(mod13)

2200= (240)5 ≡35(mod13)≡ 9(mod13)

2400=(2200)2 ≡92(mod13)≡ 3(mod13)

22000=(2400)5 ≡35(mod13) ≡9(mod13)

22005=22000.25 ≡6.9 (mod13)≡ 2(mod13)

Remainder is 2 when 22005 is divided by 13.

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q27. Find the number of digits in the number

22005 .52000 when written in full.

Ans.

22005.52000

= 25.22000.52000

= 32. 102000

Number of digits =2 +2000

=2002

Q28. Arrange the following in ascending

order: 25555, 33333, 62222.

Ans.

25555 =(25)1111 =321111

33333 =(33)1111 =271111

62222 = (62)1111=361111

Since exponents are equal therefore

bases will decide the order of

numbers Hence, Ascending order:

33333, 25555, 62222

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q29. Find all the positive perfect cubes that

divide 99.

Ans.

99= (93)3 =7293

Cubes of all factors of 729 will

divide 99

Factors are: 1,33,93,273,813,2433,

7293

Required numbers are: 1,33,36,39,

312, 315, 318 6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q31.(123456)2 +123456 +123457 is the square of…….

Ans.

Here, (123456)2 +123456 +123457

= (123456)2 +123456 +123456 +1

= (123456)2 +2*123456 *1+12

= (123456+1)2

= 1234572

Required number is 123457

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q32.How many four digit numbers can be formed

using the digits 1, 2 only so that each of

these digits is used at least once ?

Ans.

Following cases are possible:

i. 1 used thrice and 2 once

ii. 1 used twice and 2 twice

iii. 1 used once and 2 thrice

Number of four digit numbers in i and iii cases

= 4 each

Number of four digit numbers in ii case

= 6

Required number of numbers=14

Q33. Find the number of perfect cubes

between 1 and 1000009 which are exactly

divisible by 9.

Ans. Since, 1<x3 <1000009 ie. 1<x <101 Perfect cubes divisible by 9 will be cubes of multiples of 3. i.e. x is a multiple of 3

But, 101=3x33 + 2

Between1 and101 there are 33multiples of 3

Required number of perfect cubes =33

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q34. Find the number of positive integers less than or

equal to 300 that are multiples of 3 or 5, but are

not multiples of 10 or 15.

Ans. No. of multiples of 3 = [300 /3 ]=100

No. of multiples of 5 =[300 /5 ] = 60

No. of multiples of 3 and 5 both =[300 /15 ] = 20

No. of multiples of 10 = [300 /10 ]= 30

N0 of multiples of 15 =[300 /15 ] = 20

No. of multiples of 10nd 15both =[300 /30 ] =10

Therefore, Required number of numbers

= (100 +60-20) - (30+20-10)

= 140-40 = 100 Here [ X ] denotes greatest integer less than or equal

to x .

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs

Q35.The product of the digits of each of the three – digit

numbers 138, 262 and 432 is 24. Write down all

three digit numbers having 24 as the product of the

digits.

Ans.

24 can be written as a product of three numerals as –

1x 3 x8 ,1x 6 x4 ,2 x4 x3 ,2 x6x 2 .

For three different numerals there are 6 arrangements of

each possible product and for fourth product having 2

two’s number of arrangements will be 3

No of three digit numbers having product of their digits

24 is 21.

They are 138, 183, 318, 381, 813, 831, 164, 146, 461,

416, 614, 641, 243, 234, 342, 324, 432, 423, 262, 226,

622 6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs

Q36. Find the numbers lying between 60 and 70,

each of which divides 248-1.

Ans.

248-1=(26-1)(26+1)(212+1)(224+1)

212+1 and 224+1 are greater than 70

Therefore, Numbers between 60 and

70 are 26-1 and 26+1

i.e. 63 and 65

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs

Q37. A printer numbers the pages of a book starting

with 1. He uses 3189 digits in all. How many

pages does the book have?

Ans.

No. of digits used in one digit number = 9×1 = 9

No.of digits used in two digit number = 90×2 = 180

No. of digits used in 3 digit number = 900×3 = 2700

No.digits used till three digit numbers = 9 + 180 +

2700 = 2889

Remaining digits used for 4 digit numbers

= 3189 – 2889 = 300

Therefore, number of 4 digit numbers = 300/4 = 75

Number of pages = 1074

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs

Q38. Find the largest prime factor of

312 +212 – 2.66.

Ans.

312 + 212 - 2.66

= (36)2 + (26)2 – 2.36.26

= (36 – 26)2 = {(33 – 23) (33 + 23)}2

= {(3–2)(32+3.2 +22).(3+2)(32–3.2+ 22)

= {19.5.7}2

Therefore, Largest Prime Factor = 19

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q39. Find the value of

S= 12- 22 +32-42+…………-982+992

Ans.

S= 12-22+32-42+………-982+992

=(12-22 )+(32-42)+…+(972-982)+(992-1002) +1002

= ( -3-7-11…………….-199) +10000

{ n2-(n+1)2 =-(2n+1) }

= -(50/2)[ 2*3+49*4] +10000

= 4950

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q40. Find the smallest multiple of 15 such that each digit

of the multiple is either‘0’or ‘8’.

Ans. Prime factors of 15 are 3 ,5.

Therefore any multiple of 15 must be divisible

by 3 and 5.

As the required no has to be divisible by 5, it

should end in zero (the option 5 is not applicable here)

Also, the given no must be divisible by 3.

Also, we want the smallest multiple of 15.

Therefore the only possibility is 8880.

The required no is 8880.

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q41. If n is a positive integer such that n/810 =

0.d25d25… where d is a single digit in

decimal base. Find ‘n’.

Ans. Let N=.d25d25… => N=d25/999 But d25/999=n/810 => n=d25X90/111 It is possible when d=9 Hence , n=750

6/27/2015

PREPARED BY:- M. GOVINDU, Principal,Kvs

Q42. Let x be the LCM of 32002-1 and 32002+1. Find

the last digit of x.

Ans.

32002=(34)500 X 32

As 34= 81 is having unit digit is 1

hence, unit digit of 32002 is 9

Unit place digit of (32002-1) = 8

Unit place digit of (32002+1 )= 0

The units digit of the product is 0

5 & 2 are the factors of their LCM

If 10 is factor of LCM, then it’s unit digit

will be 0

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

Q43. Let f0(X)=1/(1-X) and fn(x) = f0(fn-1(x)) Where

n = 1,2,3….Calculate f2009(2009)

THAN Q &

6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs

ALL THE BEST