8th Mexico 1994 Problem A1 The sequence 1, 2, 4, 5, 7, 9 ,10, 12, 14, 16, 17, ... is formed as follows. First we take one odd number, then two even numbers, then three odd numbers, then four even numbers, and so on. Find the number in the sequence which is closest to 1994.
1993 and 1995 Solution In the first 1 + 2 + ... + (n+1) numbers we have n gaps of 1, the other gaps are all 2. Now 1 + 2 + ... + 44 = 990, so in the first 990 numbers we have 43 gaps of 1, so the 990th number is 1 + 2·989 - 43 = 1936. The next number is 1937 and we then have 44 more odd numbers which takes us past 1994. So we get 1993 and 1995 but not 1994. 8th Mexican 1994 © John Scholes jscholes@kalva.demon.co.uk 28 February 2004 Last corrected/updated 28 Feb 04