1st CIS 1992 Problem 20 Find all integers k > 1 such that for some distinct positive integers a, b, the number k a + 1 can be obtained from k b + 1 by reversing the order of its (decimal) digits.
k = 3, 3 3 +1 = 28, 3 4 +1 = 82. Solution k = 10 does not work because k a + 1 is a palindrome. If k > 10, then for a  10(k a + 1). So k b + 1 has more digits than k a + 1. So we only need to consider k = 2, 3, ... , 9. wlog a  k a k a+1 > k a (k a + 1), so k a  k b - 1 ≥ k b - k a + k b-a - 1 = (k a + 1)(k b-a - 1). So k b-a - 1 < 10. If k b-a - 1 = 9, then k b- a = 10, so k = 10, which we already know does not work. Hence k b-a - 1 < 9. But (k b + 1) - (k a + 1) = (k b-a - 1)k a . This must be divisible by 9, because k b + 1 and k a + 1 have the same digit sum and hence are the same mod 9. Hence k a must be divisible by 9, so k must be 3, 6 or 9. If k = 6 or 9, then kb-a - 1 ≥ 5, so the first digit of k a + 1 must be 1 or (k b-a - 1)(k a + 1) would have more digits than k a + 1. But that means the last digit of k b + 1 is 1 and hence the last digit of k b is 0, which is impossible. So k = 3. It is easy to check that there is a solution for k = 3. 1st CIS 1992 © John Scholes jscholes@kalva.demon.co.uk 5 December 2003 Last corrected/updated 5 Dec 03