Bunuel wrote:

\(30^{30} * 29^{29} * 28^{28}*...*2^2 * 1^1 = n\). How many zeroes does ‘n’ contain at the end of the number (to the right of the last non-zero digit)?

Enter your value.

Magoosh Answer and Explanation

Some of the zeroes are easy to get: the one’s in which the number is a multiple of 10. So 30^30 will yield 30 zeroes, 20^20 will yield 20 zeroes, and 10^10 will yield 10 zeroes, giving us a total of sixty zeroes.

But here is where things start to get trickier. Don’t forget the ‘5’s and the ‘2’s. Why? Well, as long as we can factor out a combination of a ‘5’ and a ‘2’, we have 5 x 2 = 10, meaning exactly one zero. Now before you go counting all the ‘2’ in this enormous number—which might take all day—it is better to get the number of ‘5’s’ since there are fewer and you know that however many ‘5’s you count there will be enough ‘2’s to pair it off with.

5^5 = five zeroes

15^15 = fifteen zeroes

25^25 = ….here’s the doozy. 25 = 5^2. Therefore, 25^25 = 5^50, giving us a whopping fifty zeroes.

So let’s add these up:

5 + 15 + 50 = 70

Adding these to the 60 zeroes we got from all those factors of 10 gives us 130, the answer.