How many terminating zeros are in the product of 80!?

Question

How do you do this without a calculator and what's a terminating zero? Also it's 80! = 1 * 2 * 3... 80. It's not an exclamation mark...

Answer 1

Each time you have a multiple of 5 as one of the factors, it will combine with an even number to make a multiple of 10. This will contribute 1 zero to the product.

Each time you have a multiple of 25, you can combine it with a multiple of 4 to make a multiple of 100. This will contribute an additional zero to the product.

If your number was bigger, you'd have to worry about higher powers of 5. For example, each time you have a multiple of 125, you can combine it with a multiple of 8 to make 1000. This would contribute another trailing zero to the product. Going higher... 625 x 16 = 10000, 3125 x 32 = 100,000, etc.

Anyway, all you have to worry about are multiples of 5 and 25 in your case:
80 / 5 = 16, so there are 16 multiples of 5.
80 / 25 = 3 r 5, so there are 3 multiple of 25.

The answer should be 19 trailing zeroes.

Answer 2

19... I used a TI 89 Calculator, and it gave me the exact answer... you can do this with out a calculator, but I am not sure how... the number of terminating 0's is from the point at which the number ends in only 0's... for example, the last 24 digits of 80! are... 0814080000000000000000000... so you count from the 0 next to the last 8.

Answer 3

I think the answer should have 72 zeros.