Permutations and combinations - divisibility of a factorial

Asked Jan 03 '15 at 14:00

Active Jan 03 '15 at 14:04

Viewed 79 times

Question:

Find the largest value of $n$ for which $125!$ is divisible by $6^n$

Approach: I tried to find all the numbers which were a multiple of 6. The number of times such divisors occurred should simply be the value of $n$. However, this proved to be a difficult task. Any simpler approach?

asked Jan 03 '15 at 14:00

Gummy bears

3,408

2

It is sufficient to find the highest power of $3$ dividing $125!$, which is $3^{59}$ – Peter Jan 03 '15 at 14:03
@Peter How did you come up with that though? – Gummy bears Jan 03 '15 at 14:05
1

@Gummybears: Recall this Question of yours from yesterday. – hardmath Jan 03 '15 at 14:05
@hardmath I can use the same method for any number? – Gummy bears Jan 03 '15 at 14:06
2

Just as the limiting "factor" for factors of ten is counting the fives in the factorial, here the factors of three are not so many as the factors of two. So count the scarcer prime factor, using the same "floor" fractions. – hardmath Jan 03 '15 at 14:08
So I cannot use 6, as rogerl has done? @hardmath – Gummy bears Jan 03 '15 at 14:09
@hardmath So it is only applicable for prime numbers. And that too, we have to use the larger prime factor? – Gummy bears Jan 03 '15 at 14:11
Gummybears: You should confirm your new understanding by counting the factors of 3 that @Peter claims above. The computation is rather easily organized using a calculator (applet or hand-held). – hardmath Jan 03 '15 at 14:31

Permutations and combinations - divisibility of a factorial

0 Answers0