IMO shortlist 1983/24 solution

24th IMO 1983 shortlist

Problem 24

d_n is the last non-zero digit of n! . Show that there is no k such that d_n = d_n+k for all sufficiently large n.

Solution

Let L(n) be the last non-zero digit of n. The basic idea is that if L(N) and L(N') are both even and L(NN') = L(N), then L(N') must be 6.

Note first that L(n!) is always even for n > 1 because n! has more 2 factors than 5 factors. Suppose there is a period T. For k even, we always have L(k) ≠ L(2k), so T is not 1. But if n has last digit 1, then L(n!) = L(n-1!) so if the sequence had period 2, it would also have period 1, which is impossible. So T is not 2. If n has last digit 2, then n+1 has last digit 3 and n(n+1) has last digit 6. But L(k) = L(6k) for k even, so L(n+1!) = L(n-1!). However, L(k) ≠ L(4k) for k even, so L(n+2!) ≠ L(n-1!). Hence T is not 3. So T > 3. Hence (T-1)! has last non-zero digit even.

Now take m sufficiently large that 10^m > (T-1)!. Put n = 10^m - 1. Then N' = (n+1)(n+1) ... (n+T) = 10^m(10^m+1)...(10^m+T-1) = 10^m(T-1)! mod 10^2m. So L(N') = L(T-1!) and is therefore even. By assumption L(n+T!) = L(n!) and is even. So L(N') = 6.

Now take n' = 2·10^m - 1. Again L(n'+T!) = L(n'!) and is even. We have (n'+T)!/n'! = 2·10^m(2·10^m + 1) ... (2·10^m + T-1) = 2(T-1)! mod 10^2m. So L((n'+T)!/n'!) = 2L(T-1!) = 2. But that implies L(n'+T!) ≠ L(n'!). Contradiction.

24th IMO shortlist 1983