17th Mexico 2003 Problem A1 Find all positive integers with two or more digits such that if we insert a 0 between the units and tens digits we get a multiple of the original number.
15, 18, 45, or any muliple of 10 Solution Let the number be n. Let n' be the number obtained by inserting the 0. n must also divide 10n and hence 10n - n'. If the last digit of n is d, then 10n - n' = 9d. So n must divide 9d. In particular, n must be a 2 digit number. For example if d = 9, we need a two digit number ending in 9 that divides 81. There are none. Similarly, we check d = 8 giving n = 18, d = 7 no solutions, d = 6, no solutions, d = 5 giving n = 15 or 45, d = 4 so solutions, d = 3 no solutions, d = 2 no solutions, d = 1 no solutions. Finally if d = 0, then any number works. Thanks to Marco Avila Ponce de Leon for pointing out the multiple of 10, which I managed to overlook! 17th Mexican 2003 © John Scholes jscholes@kalva.demon.co.uk 1 March 2004 Last corrected/updated 2 Mar 04