How many consecutive zeroes end 1000!, exactly?

Answer 1

Thinking about it, finding the powers of 5 in 1000! will give you the answer. So...

There are 200 numbers from 1 to 1000 that are divisible by 5. (So far we know 1000! is divisible by 5^200.)

Of these 200 numbers, 40 are divisible by 25. (Now we have 5^240 as a divisor.)

Of these 40 numbers, 8 are divisible by 125. (Now we have 5^248 as a divisor.)

Finally, 625 has a fourth multple of 5.

So, 5^249 is the largest power of 5 that divides 1000!. 2^249 can be easily shown to divide 1000!, therefore 10^249 divides 1000! From this we can conclude 1000! ends in 249 zeroes.

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(Note to Conta: You need to recount the zeroes in that number. Thanks for the link, though... it confirms my answer. :) )

Answer 2

I don't think any consumer software can give you the real answer, unless it displays the answer, which is pre-calculated in a super computer or something.

So first think how factorial 1000's (1000!) expansion should be:

1 x 2 x 3 . . .998 x 999 x 1000

It should contain at least 3 consecutive 0's at the end for sure.

But if you are really lazy, click the link below and you will find that 999! ends with 249 no. of zeros.

So 1000! should end with 252 consecutive zeros.

Answer 3

3

there's 1 to start with, then three simultaneous zeros that follow up. Meaning that, there is three zeros to end 1000.

peace and love,
rob