Math IOQM

1) P(x) = x^4- 4x^3 + 6x^2 - 4x - 2005. Find the integer part of the product of all non-real roots of (P). S1) P(x) = 0 => (x-1)^4 = 2006 => (x-1) = K.(2006)^(1/4) where K = -+1, -+i So non real roots are x = 1 +-i.(2006)^(1/4) Multiply: 1 + 2006^(1/2) 45^2 = 2025 45^2 - 44^2 = 89 So 2006^(1/2) = 44.something Answer: 45 2) First we will prove this: In the incircle trisects the median then the sides are in the ratio 5:10:13 where 10 maps to the base where median is landing. Once we do that, since BC = 20, other sides will become 10 and 26. Then apply Heron's formula. s = 28 s - a = 28 - 20 = 8 s - b = 28 - 10 = 18 s - c = 28 - 26 = 2 [ABC] = sqrt(28.18.8.2) = sqrt(64.7.2.9) = 24.sqrt(14) m + n =38 Ok coming to the proof of the first part. Using power of the point theorem: If DF is the tangent from D, then DF^2 = DY.DX = d/3 * 2d/3 = 2d^2/9 Similarly Tangent from A = AE then AE^2 = AX.AY = DF^2 => AE^2 = DF^2 But AE = s - a from incircle properties. And DF = (s-c) - a/2 = a+...

  1) Find the largest natural number 'm' such that:  (m^2) is the difference of two consecutive perfect cubes. AND  (2m + 79) is a perfect square. S1) m^2 = (n+1)^3 - n^3 = 3n^2 + 3n + 1 Since m,n are both natural numbers, RHS will grow much faster than LHS and at some point it will irreversibly exceed it. m^2 - 1 = 3n(n+1) So RHS is a multiple of 6 => m^2 = 1 mod 6 m^2 = 25,49 satisfy this but 24 = 3n(n+1) => 8 = n(n+1) no go. 48 = 3n(n+1) => 16 = n(n+1) no go. what next? So let's try some other approach. Once you complete the square in 'n', you will get: 4m^2 = 3(2n+1)^2 - 1 (2m-1)(2m+1) = 3(2n+1)^2 LHS has 2 consecutive odd numbers which are co prime. Why? gcd(2m-1,2m+1) = gcd(2,2m-1) = 1 So RHS should split cleanly into 2 factors which don't share any prime factors. Case 1: 2m-1 = a^2 2m + 1 = 3b^2 Case 2: 2m + 1 = a^2 2m - 1 = 3b^2 Case 2: 2 = a^2 - 3b^2 => a^2 = 3b^2 + 2 => a^2 mod 3 = 2 which is not possible since it can be only 0,1 Case 1...

Q1) Let A be a set of m positive integers, where m >= 1. Show that there is a non-empty subset (B) of (A) such that sum of all elements in B is divisible by (m). S1) Let's create prefix sums like this: S1 = a1 S2 = a1 + a2 ... Sm = a1 + a2 .. am If any one of S1,S2...Sm has a 0 remainder we are done. If not, then at least 2 of them have same remainder. Then: Sk - Sj = 0 mod m If k > j Then Sk - Sj = ak + a_{k-1} ... a_{j-1} So these are the required elements. Q2)  Let (A) be a subset of ({1,2,...,2n}) such that (|A|=n+1). Show that there exist (a, b in A) such that gcd(a,b)=1. Part 2: Can we also say the same thing when |A| = n? S2) There are at least 2 consecutive numbers in A. Proof by PHP: {1,2} {3,4} ... {2n-1,2n} Each number belongs to one bucket out of these 'n' buckets. If there are n+1 numbers, at least 2 of them will fall into same bucket. Proof by contradiction(not very solid): Let's say A has only non-consecutive numbers. Then it can have only n number...

 Q 1) Find the number of ways in which three numbers in A.P. can be selected from (1, 2, 3, ..., n). (For both (n) odd and even.) S1) We have to select 2 numbers of same parity and the AM will lie within. Case 1: n is even. Select 2 odd numbers: (n/2)C2 Select 2 even numbers: same Case 2: n is odd. Select 2 odd numbers: (n+1)/2C2 Select 2 even numbers: (n-1)/2C2 simplify to get: (n-1)^2/4

Q1. Does there exist a natural number “(n)” such that (n^2 + n + 1) is divisible by (1955)? If “Yes”, provide an example and if “No”, provide a valid proof. S1. n^2 + n + 1 = 0 mod 5 (if it's div by 1955, it's also div by 5) n^2 + n = 4 mod 5 Let's check all possible remainders of n with 5. n = 0 mod 5 No. n = 1 no. n = 2 4 + 2 = 1 mod 5 n = 3 9 + 3 = 2 mod 5 n = 4 16 + 4 = 0 mod 5 So nothing works. "No". Q3. Using the properties of congruences, prove that 89 | 2^44 - 1 and 97 | 2^48 - 1 S3. 2^44 - 1= 0 mod 89 (2^11 - 1)(2^11 + 1)(2^22 + 1) = 0 mod 89 Since 89 is a prime, one of these factors has to be a multiple of 89. Let's test one by one. 2^11 = 1 mod 89. 2^5.2^6.3 = 3 mod 89 96.64 = 3 mod 89 = 7.64 = 448 = 3 mod 89 H.P. Part 2: 2^48 - 1 = 0 mod 97 = (2^6 - 1)(2^6 + 1)(2^12 + 1)(2^24 + 1) First 2 factors are small. Next: 2^12 + 1 = 0 => 2^12 = -1 mod 97 = 96 mod 97 = 32.128 => 128 = 3 mod 97 which is wrong Next: 2^24 = -1 = 96 mod 97 2^19 = 3 mod 97 ...

Q1)   A square of dimension (n-1) * (n-1) is divided into (n-1)^2 unit squares in the chess board manner. Now, each of these (n^2) vertices of these squares are to be painted black or blue such that each unit square has exactly two black vertices. In how many ways can this be done? S1. Look at an individual square. There are 2 ways for it to yield 2 black vertices. One horizontal edge blue(or black), other black(or blue). Or 1 color each on top edge, and similarly on bottom edge. Once you expand the pattern, there will be only 2 ways to get it done. Each row alternates or each column alternates or both alternate. Case 1: Each row alternates. So each row can start with one of 2 colors and then the pattern is fixed. So 2^n ways since there are 'n' rows. Case 2: Each column alternates. 2^n again. Case 3: Both alternate. Exactly 2 ways like a chessboard. Total: 2^n.2 - 2 = 2^(n+1) - 2. Q2. S2. As the diagram shows a 2x2 bounding box can hold 2 different squares. That rule actually ...