5th Mexico 1991 Problem A2 n is palindromic (so it reads the same backwards as forwards, eg 15651) and n = 2 mod 3, n = 3 mod 4, n = 0 mod 5. Find the smallest such positive integer. Show that there are infinitely many such positive integers.
515 Solution To meet the congruence conditions, n must be 35 mod 60. So it has last digit 5. Hence it must have first digit 5 and hence must be at least 35 + 8·60 = 515, which works. So 515 is the smallest. Consider 51...15, where there are 3k+1 1s. It is obviously 0 mod 5 and 3 mod 4. The sum of the digits is 2 mod 3 and hence the number is 2 mod 3. It is obviously palindromic. 5th Mexican 1991 © John Scholes jscholes@kalva.demon.co.uk 28 February 2004 Last corrected/updated 28 Feb 04