MMO2025 Round 4

Question

Let the sum of the elements of a set X be denoted by S(X). How many ways can we divide the numbers 2^1, 2^2, , 2^10 into sets A and B such that the equation x^2 - S(A)x + S(B) = 0 has a positive integer solution? (The sets A or B may be empty.) *To divide the numbers c_1, c_2, , c_n into sets A and B means that A and B are disjoint sets and each of the numbers c_1, c_2, , c_n must belong to exactly one of them.*

Accepted Answer

*Answer: 2.* The numbers 2^1, 2^2, , 2^10 can be used to produce only and all even numbers from 0 to 2046. Thus, let S(A) = 2n for some integer 0 n 1023. Then S(B) = 2^1+2^2++2^10-S(A) = 2046-2n. Moreover x^2-S(A)x+S(B) = x^2-2nx+(2046-2n) = 0 implies 23 89 = 2047 = (n+1)^2 - (x-n)^2 = (2n+1-x)(x+1). The only positive integer solutions are x = 22 and x = 88 with n = 55. Hence, there are exactly two ways to divide the numbers.