3rd CaMO 1971 Problem 7 Find all five digit numbers such that the number formed by deleting the middle digit divides the original number.
Let the number be n = AhB, in other words 1000A + 100h + B, with 10 ≤ A  10, then k(100A + B) - 1000A ≥ 100A + B > 100h + B, which is not possible. If k  kB, which is not possible. So we must have k = 10. Hence 100h = 9B. But the that implies that 9 divides h, so h = 0 or 9. It cannot be 9, because 9B < 900, so the only solution is h = B = 0. Hence n = 10000, 11000, 12000, ... , or 99000. [It is obvious that they are solutions.] 3rd CaMO 1971 © John Scholes jscholes@kalva.demon.co.uk 10 June 2002