IMO 1962 Problem A1 Find the smallest natural number with 6 as the last digit, such that if the final 6 is moved to the front of the number it is multiplied by 4.
We have 4(10n+6) = 6·10 m + n, where n has m digits. So 13n + 8 = 2·10 m . Hence n = 2n' and 13n' = 10 m - 4. Dividing, we quickly find that the smallest n', m satisfying this are: n' = 7692, m = 5. Hence the answer is 153846. s are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 4th IMO 1962 © John Scholes jscholes@kalva.demon.co.uk 21 Sep 1998 Last corrected/updated 24 Sep 2003