Let A, B be two sets of N consecutive integers. If N = 2003, can we form N pairs (a, b) with a ∈ A, b ∈ B such that the sums of the pairs are N consecutive integers? What about N = 2004?
Answer
Yes, no.
Solution
wlog A = B = {1, 2, ... , N} - if we have a solution for A = {a+1, a+2, ... , a+N} and B = {b+1, b+2, ... , b+N}, then subtracting a from every element of A and b from every element of b gives a solution for A = B = {1, 2, ... , N}. Suppose the sum set is (m+1), (m+2), ... , (m+N). It has sum N(2m+N+1)/2 and A and B each have sum N(N+1)/2, so we must have 2m = N+1, hence N must be odd. So we cannot do it for N = 2004.
Suppose N = 2M+1, take the pairs (1,M+1), (3,M), (5,M-1), ... , (2M+1,1), (2,2M+1), (4, 2M), ... , (2M, M+2).
© John Scholes
jscholes@kalva.demon.co.uk
19 Oct 2002