Math IOQM

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

Q1. (combinatorial geometry problem) There is a 21-sided regular polygon with vertices (A1, A2 .... A21). Triangles are formed by joining these vertices. a) How many of these triangles are acute-angled ? b) How many of these triangles are right-angled ? c) How many of these triangles are obtuse-angled ? d) How many of these triangles are equilateral ? e) How many of these triangles are isosceles ? S1. Regular polygon can be inscribed in a circle. Each arc being of equal length. For e.g. equilateral triangle will create 3 arcs of 120 degrees each. So 21-gon will create 21 equal-sized arcs. Total possible triangles = 21C3 = 21.20.19/6 = 19.70 = 1330 b) 0 right angle triangles. Why? By Thales' theorem, diameter inscribes right angle on circumference. And a diameter divides circle into 2 equal parts. In an odd sided regular polygon, we cannot have equal number of arcs on both sides of the diameter. c) By inscribed angle theorem, for an obtuse angle on circumference, the intercept...