Math IOQM

Q1.  Find the number of triangles whose angular points are at the angular points of a given polygon of (n) sides, but none of its sides are the sides of the polygon. S1. Simple application of  Kaplansky's first lemma . Formula derived is: (n-k-1)C(k-1) + (n-k)C(k) Here k = 3 (n-4)C(2) + (n-3)C(3) = (n-4)(n-5)/2 + (n-3)(n-4)(n-5)/6 = (n-4)(n-5)/2[1 + (n-3)/3] = n(n-4)(n-5)/6

1. Is 2026 a semiprime(product of only 2 primes)? Yes 2. Is 2026 a deficient number(sum of its proper divisors is less than the number itself)? Yes 3. Is it a happy number? If you keep squaring the digits and adding and then repeating - does it reduce to 1? Yes 4. Is it sum of 2 squares? 45^2 + 1^2 5. Is 1013 sum of 2 squares? = 22^2 + 23^2 ( Fermat theorem on sum of 2 squares )

Q1. (n) is a positive integer. (A) is a set such that A={1,2,...,n}. Let (t_n) denote the number of subsets of (A) such that the arithmetic mean (AM) of the elements is an integer. Prove that (t_n) and (n) are both odd or both even. S1. 1. Let us ignore empty subset since A.M. is not defined for that. 2. Subsets of size 1: {1}, {2} ... {n}. There are 'n' such subsets. And each of them has their A.M. as integer. 3. Now let's consider the subsets with size >= 2 which have an integer A.M. Let G be the set of all such subsets. Let's consider a function 'f' defined for subsets of size >= 2. f(S) = S - {k} where k is the A.M. of S and k is present in S. f(S) = S + {k} where k is the A.M. of S and k is not present in S. For e.g. S = {1,3,8}, k = (1+3+8)/3 = 4 4 is not there in S. f(S) = {1,3,4,8} S = {1,2,3}, k = (1+2+3)/2 = 3 3 is there in S. f(S) = {1,2} Note 1: S and f(S) have the same A.M. Why? Let S = {a1,a2...,ak} AM = sigma(a_i)/k If you remove AM, the ne...

Q1. Find integer solutions for x^2 + y^2 = 2025 S1. Mod 9 of a square gives 0,1,4,7 Here RHS is 0. For LHS to be 0, both have to give mod 0. => x = 3a, y = 3b => a^2 + b^2 = 225 Again mod 9 is 0 a = 3c, b = 3d c^2 + d^2 = 25 (c,d) = (0,+-5) (-+5,0) (-+3,-+4) (-+4,-+3) Total solutions = 2 + 2 + 4 + 4 = 12 To find actual solutions multiply c,d with 9.

Theorem: If a1x1 + a2x2 + .... an.xn = K then this Diophantine equation has a solution if d | K where d = gcd(a1,a2...an). We can note the similarities with Bezout's lemma here. Practice: Which of these Diophantine equations have integer solutions? 1. 21x + 15y = 17 2. 7x + 17y = 27 Theorem: If ax + by = c and d | c, where d = gcd(a,b), then this Diophantine equation has infinitely many solutions, and the solutions are of the form x = x0 + (b/d).k, y = y0 - (a/d).k, where k is integer and (x_0, y_0) are particular solutions of (ax + by = c); and x0, y0 are integers.

Q1. Solve for x: 32x = 79 mod 1225 S1. 32x = 79 mod 25 => 7x = 4 mod 25, since 7*3 = 21 = -4 m 25 => 7*-3 = 4 m 25 => x = 22 m 25 32x = 79 mod 49 => 32x = 30 m 49 => 16x = 15 m 49, since 16.4 = 64 = 15 m 49 => x = 4 m 49 Use CRT now: x = p.25 + q.49 m 25 22 = q.-1 m 25 => q = -22 = 3 mod 25 m 49 4 = p.25 mod 49 => p = 8 mod 49, since 25.8 = 200 = 4 mod 49 x = 8.25 + 3.49 mod 1225 = 347 mod 1225