Prove that we cannot find consecutive factorials with first digits 1, 2, ... , 9.
Solution
We can write any integer n in scientific notation as n = n x 10k, where 1 ≤ n < 10 is a real number and k is a positive integer. For example, 356 = 3.56 x 102. Suppose that (N+i)! has first digit i for i = 1, 2, ... , 9. Then (N+i)! is a real number between i and i+1. So for i = 2, ... , 9, we have that N+i is a real number such that 1 < N+i = (N+i)!/(N+i-1)! < (i+1)/(i-1). So certainly N+i < 3 and hence N+i < N+i+1. Thus we must have 1 < N+2 < N+3 < ... < N+9 < 5/4.
But ab = a b unless a b > 10, in which case ab = (a b)/10. So in any case ab ≤ a b. Hence (N+4)! = (N+1)! N+2 N+3 N+4 < 2 (5/4) (5/4) (5/4) = 250/64 < 4. Contradiction (because it is supposed to have first digit 4).
≠ John Scholes
jscholes@kalva.demon.co.uk
14 Oct 2002