Trailing Zeros are the end of a number...

[Macaroni]

This problem is from the MGMAT phone app flash cards. Card #60 in the Number Properties category.

Trailing Zeros are the zeros at the end of a number (e.g., 35,400 has two Trailing Zeros, 7,000 has three Trailing Zeros, and 458 has none). How many Trailing Zerso does 15! have?

Answer: 3

The first hurdle is to determine what causes Trailing Zeros in a number. Multiplication by 10 adds one zero to the end of a number, multiplication by 100 (or two 10's) adds two zeros to the end, multiplication by 1,000 (or 3 10's) adds three zeros to the end, and so on.

15! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 13 x 15

There is a 10 in the product so that would be one Trailing Zero, but are there any other 10's? A 10 comes from any combination of a 5 and a 2; there are lots of 2's in 15!, but there are only three multiples of 5: 5, 10 (2x5), and 15 (3x5). This means that there are only three 10's in the product of 15!, so there will b three Trailing Zeros.

My Question:

Why are there only 3 10's? It looks like I would get pairs of 10 from the following:

10
2 x 5
15 - I would get 3 5's from 15, and find 2 2's from 4, and a 2 from 14

I would think that the answer is 5 Trailing Zeros?

Thank-you for your help in advance!

[RonPurewal]

Remember"”"”you're dealing with products of prime numbers.
E.g., 15 = 3 x 5. There's only one 5 there, not three of them. (You can add three 5's to get 15, but addition is irrelevant here.)

[RonPurewal]

You can also disprove your current line of thinking by considering small counterexamples. (Big counterexamples work too, though you'd waste more time multiplying them out.)
E.g., if you think, incorrectly, that 15 gives "three 5's" (and also, this time correctly, that 4 gives two 2's), then you'll conclude that 15 x 4 must end with two 0's. But 15 x 4 = 60, which clearly doesn't end with two 0's.

[Macaroni]

Oooh, okay, thank-you!

[RonPurewal]

Sure.