18th BMO 1982 Problem 3 a is a fixed odd positive integer. Find the largest positive integer n for which there are no positive integers x, y, z such that ax + (a + 1)y + (a + 2)z = n.
(a 2 +5a+4)/2 Solution Put s = x+y+z, so ax + (a+1)y + (a+2)z = as+y+2z. For fixed s, the smallest value this can achieve is as+3 and the largest is (a+2)s-3. It can also take all values in between, because (x,y,z) = (s-2-m,m+1,1) gives as+3+m for 1 ≤ m ≤ s-3 and (x,y,z) = (1,s-2-m,m+1) gives as+s+m for 1 ≤ m ≤ s-3. We get a gap between the values for s and the values for s+1 iff (a+2)s-3 < a(s+1)+2 or 2s < a+5. We are given that a is odd, so we have a gap iff s < (a+5)/2. The largest element in the last gap is thus a(a+5)/2 + 2 = (a 2 +5a+4)/2. 18th BMO 1982 © John Scholes jscholes@kalva.demon.co.uk 10 February 2004 Last corrected/updated 10 Feb 04