Puzzle I came up with last night, no calculators allowed:

## The Puzzle

The final 12 digits of N! (N Factorial) are all zeroes. What is the smallest number N can be?

## Further

Generalising this, i.e. finding a formula for the smallest factorial ending in M zeroes, is really quite a pain to do. I’ve not managed it. It would be very easy to write an algorithm for finding solutions, but that’s less fun and not in the non-calculator spirit.

The simpler problem of finding the number of trailing zeroes of any given factorial is made trivial by the formula given in the hints below.

## The Solution

SPOILERs!

### Hint 1

N! must be a multiple of 10^12

### Hint 2

Prime factors of 10 are 5 and 2.

### Hint 3

Multiples of 2 always outnumber multiples of 5. Therefore, you only need to keep track of multiples of 5.

### Hint 4

Remember that some multiples of 5 have more than one 5 in their prime factors.

### Hint 5

There’s even a formula! (You shouldn’t need to use this but it’s here)

### Answer

The solution is N=50