# What is 2 + 2 + 2 + 3 + 3 + 3 + 4

## Factorial and the ending zeros

Have you used the calculator or your mobile phone and calculated the first examples? And maybe the prime factors sorted? Then you are well prepared for the solution.

### solution

1 · 2 · 3 · 4 · 5 = 120, so has a zero at the end.

1 · 2 ·… · 9 · 10 = 3 628 800, so has two zeros at the end.

1 · 2 ·… · 14 · 15 = 1 307 674 368 000, so it has three zeros at the end.

### Where do the zeros at the end of a product come from?

If you break down every factor of a product into its prime factors, you can see how often the prime factors 2 and 5 are in a product:

Example:

10! =

1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 =

1 · 2 · 3 · 2 · 2 · 5 · 2 · 3 · 7 · 2 · 2 · 2 · 3 · 3 · 2 · 5 =

1 · 2 · 2 · 2 ·2 · 2 · 2 · 2 · 2 · 3 · 3 · 3 · 3 · 5 · 5 · 7 =

1 · 2^{8} · 3^{4} · 5^{2} · 7 =

1 · 2^{6} · 3^{4} · 2 · 5 · 2 · 5 · 7 =

1 · 2^{6} · 3^{4} · 7 · 10 · 10

Each pair of prime factors 2 and 5 result in a factor of 10 and ultimately a zero at the end of the product. In the example above there are two fives and eight twos, so that gives two pairs of factors 2 and 5. Therefore, there are two zeros at the end of the product.

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