Thank you Vassil Vassilev, Yr 11, Lawnswood High School, Leeds for the solution below, well done! Danny Ng, 16, from Milliken Mills High School, Canada sent a very similar solution and so did Koopa Koo of Boston College.

First I tried to convince myself that 100! has 24 noughts. I did that by counting the number of 5s in the numbers from 1 to 100 which are all multiplied together. I did that because a zero at the end can only be produced by multiplying an even number with a 5 and there are more even numbers than multiples of 5 in the product.

Also there is another way to find the number of zeros. This is by:

When we add this two together we get 24 which is exactly the number of noughts in 100!

So to see if my rule works I will find how many noughts are there in 1000!:

The number of zeros has to be a whole number so the number of multiples of 5 ⁴ is 1 which is the integer part of 1.6 (written [1.6] ). Note that the process stops when division by 5 gives a number less than 5. If we add those answers together we will get the number of noughts. 200 + 40 + 8 + 1 = 249. From here we see that my rule works.

So to get the number of noughts in 10 000! we just divide by 5 to get the number of 5s:

2000 + 400 + 80 + 16 + 3 = 2499

To get the number of noughts in 100 000!:

20000 + 4000 + 800 + 160 + 32 + 6 + 1= 24999

Here is how Koopa Koo gave the solution for 1 000 000!.

Let [x] denotes the greatest integer that does not exceed x.

The number of right most zeros of 1 000 000! = [1000000/5] +[1000000/5 ²] +[1000000/5 ³] +[1000000/5 ⁴] +[1000000/5 ⁵] +[1000000/5 ⁶] +[1000000/5 ⁷] + [1000000/5 ⁸] = 200000 + 40000 + 8000 + 1600 + 320 + 64 + 12 + 2 = 249998.