Math IOQM

1. Find the total number of words formed by the letters a,a,b,b,c. S1. 5!/2!.2! 2. In how many ways can you arrange 2 beads of red colour, 2 beads of blue colour and 3 beads of black colour in a ring? Note that beads with the same colour are identical and arrangements that can be obtained by rotating (or) flipping the ring are identical. S2. First attempt(wrong): (7-1)! circular permutations Divide by 2!2!3! to account for repeated elements. Divide by 2 to account for flipping. Answer = 15 This is wrong, correct answer is 18. Why? Let's see using a smaller example. 2 blue, 3 black. (5-1)! circular permutations Divide by 2!3!. So far so good. Mistake is when we divide by 2 for flipping. Why do we even divide by 2 for flipping. Let's say we have 4 distinct beads. Here is one necklace: 1 --- 2 |       | 4 -- 3 and flipping it we get: 2 --- 1 |       | 3 -- 4 So 1-2-3-4 was clockwise and it became counter clockwise. Flipping gives a distinct perm...

Q1. Does there exist an integer x satisfying the following conditions? 10x = 1 mod 21 5x = 2 mod 6 4x = 1 mod 7 S1. No. Why? [1] =>  10x = 1 mod 21 => 10x = 1 mod 3 and 10x = 1 mod 7 => x = 1 mod 3 and 3x = 1 mod 7 => 36x = 12 mod 7 => x = 5 mod 7 So we have x = 1 mod 3 and x = 5 mod 7 [2] =>  5x = 2 mod 6 => 5x = 6k + 2 => 5x = 2 mod 3 => 2x = 2 mod 3 => 4x = 4 mod 3 => x = 1 mod 3 And 5x = 0 mod 2 => x = 0 mod 2 So we have x = 1 mod 3 x = 5 mod 7 x = 0 mod 2 [3] => 4x = 1 mod 7 => 8x = 2 mod 7 => x = 2 mod 7 Which is a contradiction to existing solutions. Q2. For how many n between 1 and 2520 do we have n^3 = 1 mod 2520? S2. 2520 = 2^3 * 3^2 * 5 * 7 So n^3 = 1 mod 8 => n = 1 mod 8 n^3 = 1 mod 9 => n = 1,4,7 mod 9 n^3 = 1 mod 5 => n = 1 mod 5 n^3 = 1 mod 7 => n = 1,2,4 mod 7 For each of those options there is a unique solution (using CRT) between 1 and 2520. So total 9 valid solutions. Q3. Find positive integers a,b,c ...

 Homework multiple choice: (done) Q1. True/False For a polynomial f(x) with integer coefficients and integers a,b,l  we have a = b mod l => f(a) = f(b) mod l? S1. True Q2. The 2-digit integers from 19 to 92 are written consecutively to form the integer  N = 192021⋯9192. Suppose that 3^k is the highest power of 3 that is a factor of N. What is k? S2. Method 1: Since 10^m = 1 mod 9 for any m => 192021⋯9192 = 19 + 20 .. 92 mod 9 = 74/2[19 + 92] = 37*111 mod 3 = 0  mod 9 = 1 * 3 = 3 So k = 1. Method 2: Sum of digits =  (2+3+4+5+6+7+8)*10 + (1+2... 9)*7 + sum_digits(19 + 90 + 91 + 92) = 35*10 + 45*7 + 40 Doing mod 3 gives 2 + 0 + 1 = 0 mod 3 Doing mod 9 gives 8 + 0 + 4 = 3 mod 9 So 3 divides it but 9 doesn't. k = 1 = answer. Q3. Remainder of 3^89*7^86 mod 17? S3. Method 1: = 21^86*27 21 mod 17 = 4 21^2 mod 17 = 16 = -1 (-1)^43 = -1 => -1*27 = -27 mod 17 = -10 mod 17 = 7 mod 17 Method 2: 3^4 = 81 = -4 m 17 3^8 = 16 m 17 = -1 3^89 = 3^88.3 = (-1)^11.3 mod 17 =...

Q1. If x^n + y^n is a prime number for some x,y > 1 then n is necessarily: Options: Power of 2, Multiple of 4, Multiple of 3, Odd S1. Answer: Power of 2. Proof by contradiction: Assume that n has an odd factor k > 1. n = m.k x^n + y^n = (x^m)^k + (y^m)^k A^k + B^k = (A+B)(A^k-1 -A^k-2.B ....+ B^k-1) for odd 'k'. Here A = x^m and B = y^m x,y > 1 m>=1 => A+B >= 4 And the other factor is also more than 1 since A^k + B^k > A + B since k > 1. So A^k + B^k is not a prime. So n can't have any odd factor. So n has to be power of 2. H.P. For e.g. n = 2 x = 2, y = 3 4 + 9 = 13 Q2. A natural number has last digit 8 in base 16 representation. What will be the remainder, when it is divided by 8 , in decimal system S2. 0 Q3. If a prime can be written as sum of two perfect square, then the remainder when it is divided by 8 cannot be S3. Any square upon division by 8 will leave 0,1,4 as remainders. So adding 2 of them can only give 0,1,2,4,5 as remainders. And if th...

Q2. For any arbitrary integer a, 3 | a (2a^2 + 7) True or false S2. a(2a^2 + 7) mod 3 = a(2a^2 + 1) = 2a^3 + a Case 1: a = 0 mod 3 => True Case 2: a = 1 mod 3 => 2 + 1 True Case 3: a = 2 mod 3 => 4 + 2 = 6 True True = answer Q3. If a,b are both odd integers then a^4 + b^4 - 2 | 16 True or false. S3. For an even integer a^4 mod 16 = 0. Clearly. For an odd integer a^4 mod 16 = 1. Why? a = 2m + 1 a^2 = 4m^2 + 4m + 1 = 4m(m+1) + 1 m(m+1) is even = 2k => a^2 = 8k + 1 a^4 = 64k^2 + 16k + 1 H.P. So a^4 + b^4 - 2 = 0 mod 16 H.P. True Q4. If a is an odd integer then 24 | a ( a^2 - 1) S4. True Method 1: a(a^2 - 1) = (a-1).a.(a+1) In 3 consecutive integers, one is always div by 3. (a-1) and (a+1) are 2 consecutive even integers. In 2 consecutive even integers, one is div by 2, other by 4. So in total, div by 8. So div by both 8,3 H.P. Method 2: a(a^2 - 1) = (a-1).a.(a+1) = (2m)(2m+1)(2m+2) = 4m(2m+1)(m+1) = 4m(m+1)(2m+1) We need to show that m(m+1)(2m+1) is divisible by 6. Clearly m(m...

Q1. For any n>=1 prove that 1/1^2 + 1/2^2 ... 1/n^2 <= 2 - 1/n S1.Using induction: we need to show that 2 - 1/n + 1/(n+1)^2 <= 2 - 1/(n+1) Rearrange to get: 1/(k+1)^2 <= 1/k(k+1) H.P.

Q1. On a board, the numbers a, b, c, d, e, f are written in a circle (say counter-clockwise). You may increase two neighbouring numbers by 1. For which case is it impossible to equalize all of them by a sequence of such steps? Select any one option a = 4, b = 10, c = 5, d = 6, e = 9, f = 2 a = 3, b = 9, c = 7, d = 6, e = 9, f = 4 a = 4, b = 11, c = 7, d = 7, e = 10, f = 2 a = 7, b = 10, c = 5, d = 8, e = 9, f = 3 S1. First a partially correct approach which works in this case but will not work always. Final sum of all numbers is even. All the increments are also even.(multiple of 2) So the original sum of all the numbers should also be even. Only option 3 fails. Hence that's the answer. But that approach fails for: a=2,b=4,c=2,d=4,e=2,f=4 So the better way to solve it is this: Let there be 2 groups:  Group 1: a,c,e Group 2: b,d,f Each time we increment, we increment by 1 in each group. So the original sum in both the groups should be same. Else it won't work. Q2. On a 10x10 boa...

Q1. Let O be the Circumcentre of triangle ABC. Find the value of  angle BAC + angle OBC in degrees. S1. Angle OBC = angle OCB since OB = OC in triangle OBC. 2OBC = OBC + OCB = 180 - 2A => OBC = 90 - A angle BAC = A OBC + BAC = 90 deg = answer Q2. Triangle ( AMC ) is isosceles with ( AM = AC ). Medians ( MV ) and ( CU ) are perpendicular to each other, and ( MV = CU = 12 ). What is the area of ( \triangle AMC )? S2. Let the centroid be G. It will divide the triangle into 3 smaller ones with same areas => GCA,GMC,GMA Area of MGC is easy to find since GM = 2/3*12 = 8 = CG and it is a right triangle. [MGC] = 1/2 * 8 * 8 = 32 [AMC] = 3*32 = 96 = Answer Q3. In a ( \triangle ABC ), let ( O, G, H ) denote its circumcenter, centroid, and orthocenter respectively. If GH/OH = m/n, where ( m, n ) are relatively prime positive integers, find ( m + n ). S3. Assume that ABC is not equilateral else it will become 0/0. Since this question doesn't mention the type of triangle we can use a quic...