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/sci/ - Science & Math

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6220606 No.6220606 [Reply] [Original]

What is the smallest value of n such that n! (n factorial) is a multiple of a googol (10^100)?

>> No.6220616

70?

>> No.6220646

This is easier if you think of numbers as collections of powers of prime factors rather than counts. To be a multiple of a googol you need 100 factors of 2 and 100 factors of 5 (making 100 factors of 10, but 10 isn't prime).

As you multiply numbers together to get a factorial you'll be accumulating factors of 2 faster than you accumulate factors of 5, so the first one with at least 100 factors of 5 should be your answer. So obviously 500! will have at least that many, because you'll have included 100 numbers which are multiples of 5. But many of those numbers contain *several* factors of 5 (e.g., 25 has two, 125 has three) so you'll reach the necessary 100 factors of 5 a bit before 500!.


>>6220616

70! is greater than 10^100 but it isn't a multiple of 10^100 (can't post the actual value of 70! because the spam filter trips, wtf?)

>> No.6220658

>>6220606
I don't know what a googol is, but your answer is 405 (by brute force).

>> No.6220669 [DELETED] 

>>6220658
>brute force
Just in case anybody wants to see what the brute force looks like (and code tags are working):

nmin_ t n nfac | nfac `rem` t == 0 = n
| otherwise = nmin_ t n' nfac' where n' = n + 1
nfac' = n' * nfac
nmin t = nmin_ t 0 1

(in Haskell), with nmin 100^10 = 405 answering the question.

>> No.6220688

>>6220646
here is wisdom

>> No.6220841
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6220841

>>6220606
pic related.

just count the numbers <500 which can be described like
X=P*5^k, k>=2, wher P is not a multiple of 5.

look for k=2, 3, 4 , infer the pattern

Concluder to 95, so 500-95=405 is the final answer

>> No.6220884

>>6220646
OP here, this is exactly the way I was thinking.

I found the answer a couple of hours ago, but it's nice to see that it's right.

Thank you!

>> No.6221484

>>6220658
Googolplexianthicalionaelx.
Graham's cubical shit is nothing compared to it.