## June 1999

Let the number of medals remaining before day N = X.
From the way that medals are passed out, for _{N}L=1,2,...,N–1 we have
X. We solve this recurrence to find
_{L+1} = (6/7)(X_{L} – L)
Then the initial number of medals must be
Because this is an integer, we must have that 7. Since ^{N} (N–6)6 and 7 are relatively prime, we conclude
that6 is a divisor of ^{N–1}N–6.
We claim that N 6, because
if N > 6 then 6, so that this divisibility
is impossible. Checking the integers ^{N–1} > N–6 > 0N=2,3,4,5,6, we find that
the only solution is N=6 and X.
_{1}=36We started with 1+5=6 medals were handed out leaving X.
On the second day, _{2}=302+4=6 medals were handed out leaving X.
Similarly _{3}=24X, _{4}=18X, and _{5}=12X, so that on the last (sixth)
day, the remaining _{6}=6N=6 medals were awarded. |

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