Ponder This

Let the number of medals remaining before day L be written as X_L. We have N = X_N. From the way that medals are passed out, for L=1,2,...,N–1 we have X_L+1 = (6/7)(X_L – L). We solve this recurrence to find

Then the initial number of medals must be

Because this is an integer, we must have that 6^N–1 is a divisor of 7^N (N–6). Since 6 and 7 are relatively prime, we conclude that
6^N–1 is a divisor of N–6. We claim that N 6, because if N > 6 then 6^N–1 > N–6 > 0, so that this divisibility is impossible. Checking the integers N=2,3,4,5,6, we find that the only solution is N=6 and X₁=36.

We started with X₁=36. On the first day, 1+5=6 medals were handed out leaving X₂=30. On the second day, 2+4=6 medals were handed out leaving X₃=24. Similarly X₄=18, X₅=12, and X₆=6, so that on the last (sixth) day, the remaining N=6 medals were awarded.