Maths Olympiad Solution

7/27/2019 Maths Olympiad Solution

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SOLUTIONS

Exercise 1.1

Ans 1) Since 41 is prime and (5,41)=1 , by Fermats Theorem

)41(mod1540 , Hence 52039=5

40.50+39=(540)50.539=150.539(mod41)=539(mod41)

)41(mod162552

= , )41(mod1021654 = , )41mod1810058 = ,

)41(mod4516

, )41(mod16532 ,

)41(mod5).16.(10.165.5.5.55 243239 =

)41(mod33

33)41(mod52039

=

Ans 2) We can write ))10(mod(010 ++ nn

))10(mod(10 + nn , ))10(mod()10( 33 + nn

= -1000(mod(n+100)

))10)(mod(1001000(1003

+++ nn

))10(mod(900 + n

Ie. n+10 should divide -900. The largest such N is 900-10=890. As n+10

cnnot be greater than 900900 = and the greatest devisor of

900900 is

So the largest +ve integer n, such that n3+100 is divisible by n+10 is

890

Ans 3) 2

5

=32)11)(mod1(

, 2

55

=(2

5

)

11 )11(mod)1( 11

)11(mod1 . This implies

)11(mod01255

+

Hence it is divisible by 11.


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Ans 4)We shall group the terms as follows

(11997+19961997)+ (21997+19951997)+ (31997+19941997)+..+

(9981997+9991997)

AS a 2n+1 + b 2n+1 is divisible by a+b, Value in Each bracket is divisible

by 1997, Hence it is divisible by 1997

Ans5.(a) 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

Ans (b) 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

Ans (c) 55% of x=240 => x=240*100/55

77%of x=(77*240*100)/55*100=336

Ans1(d) Flowers in basket become double after every minute.

Therefore, Basket was half full 1 minute before it becomes full.

i.e. in 9 minutes.

Ans 6). Let, Digit at Hundreds place=x , Digit at Tens place=y ,Then

Digit at ones place=2y. Also sum of digits=7=> x+y+2y=7 => 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


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Hence, Number=124

7(a). 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.

Ans7(b). 3 5 9 7

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

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

{Solve and justify answers yourself, Explanation required in

Examination}Ans 8):-

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

Ans) 9). a) 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


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Greatest four digit multiple of 2,3,4,5,6,7 =420x23=9660

Required number=9660+1=9661

Ans 9b) 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.

Ans 9c) 234,23456,2345678 are even.

2345 is divisible by 5 and 234567 is divisible by 3.

Hence, 23456789 is prime

Ans. 10) 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

Ans 11) 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

Ans 12) 3111-1714=1711((31/17)11-173 )

=1711((31/17)11-4913 )

But 1


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3111-1714


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62222 = (62)1111=361111

Since exponents are equal therefore bases will decide the order of

numbers Hence, Ascending order: 33333, 25555, 62222

Ans 19) 99= (93)3 =7293

Cubes of all factors of 729 will divide 99

Factors are= 1, 33, 93, 273, 813, 2433 , 7293

= 1,33,36,39, 312, 315, 318

Ans 20) 143 =11.958 app

100(12-143) =100(12- 11.958)

=100x.042

= 4.2

Nearest integer =4

Ans 21) Here (123456)2 +123456 +123457

= (123456)2 +123456 +123456 +1

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

= (123456+1)2

= 1234572

Required number is 123457

Ans 22) Following cases are possible:

i. 1 used thrice and 2 once

ii. 1 used twice and 2 twice

iii. 1 used once and 2 thriceNumber of four digit numbers in i and iii case

= 4 each ( one different digit can be placed at any of the four

places)

Number of four digit numbers in ii case


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= (4*3*2*1)/2*2

= 6 ( 4 digits can be arranged in 24 ways(Nr) but 1 and 2 occur

twice so actual number will be half for each)

Required number of numbers=14Ans 23) LCM of 2,3,4,5,6,and 7=420

Largest four digit number which is a multiple of 420= 9999-339

=9660

Therefore required number = 9660-1 =9659

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

Ans 24) 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 =27Ans 25) Perfect cubes divisible by 9 will be cubes of multiples of 3.

Since, 1


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But, 101=3x33 + 2

Between1 and101 there are 33multiples of 3

Required number of perfect cubes =33

Ans 26)No. of multiples of 3 = [300 /3 ]=100

No. of multiples of 5 =[300 /3 ] = 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 [ ] denotes greatest integer less than or equal to x .

Ans 27) 24 can be written as a product of three numerals as

1x 3 x 8 , 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 twos 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

Ans 28 a)

22005.52000 = 25.22000.52000 = 32. 102000Number of digits =2 +2000

=2002

Ans 28 b) 22005=22000.25

25=6(mod 13)


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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.

Ans 29) 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

Ans 30) 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.

Ans: 31) 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 = 31531+ 234.

Greatest number of 4 digits which is a multiple of 315 = 9999-

234=9765


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Therefore, required number = 9765-2= 9763

Ans:32)

No. of digits used in 1 digit number = 91 = 9

No.of digits used in 2 digit number = 902 = 180No. of digits used in 3 digit number = 9003 = 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

Ans: 33)

312 + 212 - 2.66

= (36)2 + (26)2 2.36.26

= (36 26)2 = {(33 23) (33 + 23)}2

= {(32)(32+3.2 +22).(3+2)(323.2+ 22) = {19.5.7}2

Therefore, Largest Prime Factor = 19

Ans 34)

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

= 4950Ans 35) 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)


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Also, the given no must be divisible by 3.

Therefore if you put one 8 or two eights or one 8 and zero before

zero

i.e. 80 or 880or800or8080 are not divisible by 3.Also, we want the smallest multiple of 15

Therefore the only possibility is 8880. The required no is 8880.

Ans 36) 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.

Ans 37) 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=6Ans 38) Let,N=.d25d25

Solving we get N = d25/999 , But d25/999 = n/810

Now, 925 = 37 X 25 , n/30 = 37 X 25/37

n = 25 X 30 , n = 750


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Ans 39):

32002=(34)500 X 32

As 34= 81 = (Unit digit 1) X 9

= unit digit of 32002 is 9Unit place digit of (32002-1) = 8

Unit place digit of (32002+1 )= 0

5 & 2 are the factors of their LCM

Factor of LCM must be 2X5 = 10

If 10 is factor of LCM, then its unit digit will be 0

Ans40): fn(x) = 1/ 1-fn-1(x)

f1(x) = f0(f0(x)) ie.f1(x) = (x-1)/x

f2(x)=x and f3(x) = 1/(1-x) ie. f3(x) = f0(x)

Similarly

fo(x) = f3(x) = f6(x) = f2007(x) = 1/(1-x)

f2008(x) = 1/(1- f2007(x)) , f2008(x) = 1/1-(1/(1-x))

= (x 1)/x

f2009(x) = 1/1-((x-1)/x) = x

f2009 (2009) = 2009

NOTE:

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Maths Olympiad Solution