Find the number of zeroes at the end of (51! -50!)

By legendre's factorial formula (which is quite intuitive if you take time to think about it), the number of zeroes in [tex]50![/tex] is [tex]\lfloor \frac{50}{5} \rfloor + \lfloor \frac{50}{25} \rfloor = 10+2=12[/tex] zeroes.

Notice that 50 = 5^2 * 2, so it will contribute 2 more zeroes to the number.

Therefore, there will be 12+2 = 14 zeroes in total.

Rough explanation of why the above works:

Note that multiplying anything by 10 adds one zero at the end of a number, and 10=2*5.

In the prime factorization of a factorial function, the power of 5 will obviously be way smaller than the power of 2 (you could calculate exactly but i'm not going to do that here).

In this case, we need to find the power of 5's in the prime factorization of 50! * 50, which was demonstrated above.

You could find the power of 2's to absolutely confirm that it will work, but if you want to save time in a competition that will probably be irrelevant.