25th Swedish 1985 Problem 2 Find the smallest positive integer n such that if the first digit is moved to become the last digit, then the new number is 7n/2.
153846 Solution Put n = 10 m a+b, where 1 ≤ a ≤ 9. Then 7n/2 = 10b + a, so 13b = a(7·10 m -2). Obviously 13 cannot divide a, so it must divide 7·10 m -2. Hence 10 m = 4 mod 13. We have 10 1 = 10, 10 2 = 9, 10 3 = -1, 10 4 = 3, 10 5 = 4 mod 13. So the smallest possible m is 5. Obviously the smallest possible a is 1, and we get b = 53846. Check: (7/2)153846 = 538461. 25th Swedish 1985 © John Scholes jscholes@kalva.demon.co.uk 16 February 2004 Last corrected/updated 16 Feb 04