



 
This is a fairly subtle problem. In most cases, finding the last nonzero digit of (n+1)! from n and the last nonzero digit of n! is straightforward. It would be very easy to find the last digits of the product of the natural numbers up to N that were not divisible by 5. After N=2 they loop with a period of 8 [omitting values of N as well that end in 0 or 5] EXERCISE: prove this. This can be extended to include factors of 10, 100, etc provided the last nonzero digit is not 5. EXERCISE: how? However, when we multiply by something ending in 5, 50, 500, etc we must be careful  note that 14 x 5 = 70 while 24 x 5 = 120. Multiplying by numbers ending in 25, 125... is even trickier. Here I pass the problem back to you. Good Hunting!  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 