In a sports contest a total of m medals were awarded over n days. On the first day one medal and 1/7 of the remaining medals were awarded. On the second day two medals and 1/7 of the remaining medals were awarded, and so on. On the last day, the remaining n medals were awarded. How many medals were awarded, and over how many days?
Solution
Let the number of medals remaining at the start of day r be mr. Then m1 = m, and 6(mk - k)/7 = mk+1 for k < n with mn = n.
After a little rearrangement, we find that m = 1 + 2(7/6) + 3(7/6)2 + ... + n(7/6)n-1. Summing, we get m = 36(1 - (n + 1)(7/6)n + n (7/6)n+1) = 36 + (n - 6)7n/6n-1. 6 and 7 are coprime, so 6n-1 must divide n - 6. But 6n-1 > n - 6, so n = 6 and m = 36.
Solutions 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.
© John Scholes
jscholes@kalva.demon.co.uk
29 Sep 1998
Last corrected/updated 29 Sep 1998