14th Swedish 1974 Problem 3 Let a 1 = 1, a 2 = 2 a 1 , a 3 = 3 a 2 , a 4 = 4 a 3 , ... , a 9 = 9 a 8 . Find the last two digits of a 9 .
21 Solution Working mod 100, we find 9 1 = 9, 9 2 = 81, 9 3 = 29, 9 4 = 61, 9 5 = 49, 9 6 = 41, 9 7 = 69, 9 8 = 21, 9 9 = 89, 9 10 = 1, 9 11 = 9, so periodic period 10. Thus we only need to find the last digit of a 8 . Working mod 10 we find 8 1 = 8, 8 2 = 4, 8 3 = 2, 8 4 = 6, 8 5 = 8, 8 6 = 4, so periodic period 4. Thus we need to find a 7 mod 4. We find 7 1 = 3, 7 2 = 1, 7 3 = 3, so periodic period 2. Now a6 is obviously even, so a 7 = 1 mod 4. Hence a 8 = 8 mod 10. Hence a 9 = 21 mod 100. 14th Swedish 1974 © John Scholes jscholes@kalva.demon.co.uk 13 February 2004 Last corrected/updated 13 Feb 04