MATHS OLYMPIAD BY M GOVINDU VICE PRINCIPAL KV

MATHS OLYMPIAD BY M. GOVINDU VICE PRINCIPAL KV, BOLARUM 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 1 (a). Find the difference between the largest and the smallest numbers that can be formed with six digits. Ans 1. (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 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

1(b) The average of nine consecutive natural numbers is 81. Find the largest of these numbers. Ans (b) Let numbers be x. x+1, x+2, x+3, x+4, x+5, x+6, x+7, x+8 Average=81 =>9 x+36/9=81 => x+4=81 =>x=77 Hence Largest number =85 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

1(c) What will be 77% of a number whose 55% is 240? Ans (c) 55% of x=240 => x=240*100/55 77%of x=(77*240*100)/55*100=336 1(d) 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? 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Ans 1(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. Q 2. 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. 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Ans 2). Let, Digit at Hundred’s place=x Digit at Ten’s place=y Then Digit at one’s place=2 y. Also sum of digits=7 => x+y+2 y=7 5 => x=7 -3 y Number=100(7 -3 y)+10 y+2 y=700 -288 y On reversing digits new Number=100(2 y)+10 y+(7 -3 y)=207 y+7 Since on adding 297 digits are reversed Therefore, 700 -288 y+297=207 y+7 => 495 y=990 , y=2 Hence, Number=124 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 3 (a) When an integer ‘n’ is divided by 1995. The remainder is 75. What is the remainder when ‘n’ is divided by 57? 3(a). Let, The Number be n, when divided by 1995 leaves remainder 75. => n=1995 xq+75 => n=57 x 35 q+57+18 => n=57(35 q+1) +18. Hence, remainder will be 18 when the number is divided by 57. 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 3(b). Find the missing digits in the following multiplication sum: 3597 *** ---------------****** * * *** ----------------****541 ----------------- Ans 3(b). 3597 753 ---------------10791 17985 25179 ----------------2708541 PREPARED BY: - M. GOVINDU, PGT(MATHS) ----------------9/18/2020 {Solve and justify answers yourself, Explanation required in

Q 4. )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)(34 -32+1) =3. 4. 8. 91. 10. 73 =26. 3. 5. 7. 13. 73 Largest prime factor of 314+313 -12 =73 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 5) a)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) Q 5. 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=420 x 23+340 Greatest four digit multiple of 2, 3, 4, 5, 6, 7 =420 x 23=9660 Required number=9660+1=9661 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

5 b)How many prime numbers between 10 and 99 remain prime when the order of their digits is reversed? Ans 5 b) 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. 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 5 c) Exactly one of the numbers 234, 23456, 2345678, 23456789 is a prime. Which one must it be? Ans 5 c) 234, 2345678 are even. 2345 is divisible by 5 234567 is divisible by 3. Hence, 23456789 is prime. 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 6) A two-digit number is such that if a decimal point is placed between its two digits, the resulting number is onequarter of the sum of two digits. What is the original number? Ans. 6) Let, Number=10 x+y If decimal is placed- (x+y)/10=1/4(x+y) y=5 x Only possible value for x is 1 Therfore, Number=15 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 7) 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 7) Let number be x X=56 a+48=72 b+64=84 c+72=96 d+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 =2016 X 49+1215 Largest multiple of 5 digits = 99999 -1215= 98784 9/18/2020 Required number = 98784 -8= 98776 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 8) Which is greater: ? 31 11 or 17 14 Ans 8) 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 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 9) Show that is exactly divisible by 199+299+399 +499+599 is exactly divisible by 5 Ans 9) 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 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 10) Find the number of perfect cubes between 1 and 1000001 which are exactly divisible by 7 Ans 10) 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, requied number of such cubes = 14 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 11) How many numbers from 1 to 50 are divisible by neither 5 nor 7, and have neither 5 nor 7 as a digit. Ans 11) 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 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 12) The square of a number of two digits is four times the number obtained by reversing its digits. Find the number. Ans 12) Let Number be 10 x+y (10 x+y)2 = 4. (10 y+x) Number is even and 10 y+x IS A PERFECT SQUARE. Possible values=25, 49, 64 and 81 Square root of 4(10 y+x) may be 10, 14, 16 , 18 10 , 14 and 16 does not satisfy other conditions Required number is 18 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 13) Find the sum of the digites in 2 2000. 52004 Ans 13) 22000. 52004 = 54. 22000. 52000 = 625. 102000 Therefore , Sum of digits =6+2+5=13 Q 14) Arrange the following in ascending order: 25555, 33333, 62222. 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Ans 14) 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 Q 15) Find all the positive perfect cubes that divide 99. 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Ans 15) 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 Q 16) Find all the integers closest to 100(12 -√ 143) Ans 16) √ 143 =11. 958 app 100(12 -√ 143) =100(12 - 11. 958) =100 x. 042 = 4. 2 Nearest integer =4 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 17 (123456)2 +123456 +123457 is the square of ………… Ans 17) Here (123456)2 +123456 +123457 = (123456)2 +123456 +1 = (123456)2 +2*123456 *1+12 = (123456+1)2 = 1234572 Required number is 123457 Q No. 18 How many four digit numbers can be formed using the digits 1, 2 only so that each of these digits is PREPARED BY: - M. GOVINDU, 9/18/2020 PGT(MATHS) used at least once ?

Ans 18) 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 case = 4 each ( one different digit can be placed at any of the four places) Number of four digit numbers in ii case = (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=14 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 19 Find the greatest number of four digits which when increased by 1 is exactly divisible by 2, 3 , 4, 5, 6 and 7 Ans 19) 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) 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No 20 Find the last two (ten’s and unit’s) digit of (2003)2003 Ans 20) 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(mod 100) (2003)2000 =(200340)50 =150 (mod 100) =1(mod 100) (2003)2003 =20032000. 20032. 20031(mod 100) =1. 9. 3(mod 100) =27(mod 100) PREPARED BY: - M. GOVINDU, 9/18/2020 PGT(MATHS) Last two digits of 2003 =27

Q No 21. Find the number of perfect cubes between 1 and 1000009 which are exactly divisible by 9. Ans 21) Perfect cubes divisible by 9 will be cubes of multiples of 3. Since, 1<x 3 <1000009 ie. 1<x <101 Also x is a multiple of 3 But, 101=3 x 33 + 2 Between 1 and 101 there are 33 multiples of 3 Required number of perfect cubes =33 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No 22 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 22) 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 N 0 of multiples of 15 =[300 /15 ] = 20 No. of multiples of 10 nd 15 both =[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. 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No 23 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 23) 24 can be written as a product of three numerals as – 1 x 3 x 8 , 1 x 6 x 4 , 2 x 4 x 3 , 2 x 6 x 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, PREPARED BY: - M. GOVINDU, 6229/18/2020 PGT(MATHS)

Q 24). (a) Find the number of digits in the number 22005. 52000 when written in full. Ans 24 a) 22005. 52000 = 25. 22000. 52000 = 32. 102000 Number of digits =2 +2000 =2002 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q 24(b) Find the remainder when 22005 is divided by 13. Ans 24 b) 22005=22000. 25 25=6(mod 13) 210= (25)2 =62(mod 13) =10(mod 13) 220= (210)2 =102(mod 13)= 9(mod 13) 240= (220)2 =92(mod 13) =3(mod 13) 2200= (240)5 =35(mod 13)= 9(mod 13) 2400=(2200)2 =92(mod 13)= 3(mod 13) 22000=(2400)5 =35(mod 13) =9(mod 13) 22005=22000. 25 =6. 9 (mod 13)= 2(mod 13) Remainder is 2 when 22005 is divided by 13. 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No 25) Find two numbers both lying between 60 and 70, each of which divides 248 -1. Ans 25) 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 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 26 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 26) Let, X= 7 q 1+6 = 7(q 1+1)-1 X= 11 q 2+10 = 11(q 2+1)-1 X= 13 q 3+12 = 13(q 3+1)-1 Hence number is 1 less than common multiple of 7, 11 and 13 LCM of 7, 11 and 13=1001 Hence, X=1001 q-1 =1001(q-1)+1000 Therefore when X is divided by 1001 will leave remainder 1000. 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 27) Find the greatest number of four digits which when divided by 3, 5, 7, 9 leaves remainders 1, 3, 5, 7 respectively. Ans: 27) Let, The Number = 3 x + 1 = 5 y + 3 = 7 z + 5 = 9 u + 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 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 28) 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: 28) No. of digits used in 1 digit number = 9× 1 = 9 No. of digits used in 2 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 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 29 Without actually calculating, find which is greater: 3111 or 1714. Ans 29) 3111 < 3211 => 3111<(25)11 => 3111<255 AND 1714>1614 => 1714>(24)14 => 1714>256 Hence 3111<255<256<1714 => 3111<1714 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 30) Find the largest prime factor of 312 +212 – 2. 66. Ans: 30) 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 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 31) Find the value of S= 12 - 22 +32 -42+…………-982+992 Ans 31) S= 12 -22+32 -42+………-982+992 =(12 -22 )+(32 -42)+…+(972 -982)+(992 -1002) +1002 = ( -3 -7 -11……………. -199) +10000 { n 2 -(n+1)2 =-(2 n+1) } = -50/2[ 2*3+49*4] +10000 = 4950 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 32) Find the smallest multiple of 15 such that of the multiple is either‘ 0’or ‘ 8’. each digit Ans 32) 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. Therefore if you put one 8 or two eights or one 8 and zero before zero i. e. 80 or 880 or 8080 are not divisible by 3. Also, we want the smallest multiple of 15 Therefore the only possibility is 8880. The required no is 8880. 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 33) A number ‘X’ leaves the same remainder while dividing 5814, 5430, 5958. What is the largest possible value of ‘X’. Ans 33) According to the given condition, 5814 = a. X + r, 5430 = b. X + r, 5958 = c. X +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. 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 34) Consider the following multiplication in decimal notations (999)x(abc) = def 132, determine the digits a, b, c, d, e, f. Ans 34) 999 x abc = def 132 LHS = (1000 – 1) abc = abc 000 -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 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 35) If n is a positive integer such that n/810 = 0. d 25… where d is a single digit in decimal base. Find ‘n’. Ans 35) Let, N=. d 25… Solving we get N = d 25/999 But d 25/999 = n/810 Now, 925 = 37 X 25 n/30 = 37 X 25/37 n = 25 X 30 n = 750 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 36 Let x be the LCM of 32002 -1 and 32002+1. Find the last digit of x. Ans 36): 32002=(34)500 X 32 As 34= 81 = (Unit digit 1) X 9 = unit digit of 32002 is 9 Unit 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 2 X 5 = 10 If 10 is factor of LCM, then it’s unit digit will be 0 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

Q No. 37) Let f 0(X)=1/(1 -X) and fn(x) = f 0(fn-1(x)) Where n = 1, 2, 3…. Calculate f 2009(2009) Ans 37): fn(x) = 1/ 1 -fn-1(x) f 1(x) = f 0(x)) ie. f 1(x) = (x-1)/x f 2(x)=x and f 3(x) = 1/(1 -x) ie. f 3(x) = f 0(x) Similarly fo(x) = f 3(x) = f 6(x) = ……f 2007(x) = 1/(1 -x) f 2008(x) = 1/(1 - f 2007(x)) , f 2008(x) = 1/1 -(1/(1 -x)) = (x – 1)/x f 2009(x) = 1/1 -((x-1)/x) = x f 2009 (2009) = 2009 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)

THAN Q 9/18/2020 PREPARED BY: - M. GOVINDU, PGT(MATHS)