Brandon continued.

I did the following:
If I factorize the number 3335, I get 3335 = 5.23.29. Now I need to know how many times the number 29 is included in 3336!. I think I can calculate that by doing the division 3335:29 = 115
My doubt is if 115 is the answer because I need to have the number 3335 115 times so the factor 29 (the greater) is contained in 3336! 115 times

Hi Brandon,

I want to look at a similar but smaller question.

What's the greater positive integer number k such that $180^k$ is a divisor of $181!$?

Analogous to what you did I wrote 180 in terms if its prime factors, $180 = 2^2 \times 3^2 \times 5$ and 180 divided by 5 is 36. You can list the multiples of 5 that divide 181!, they are

$5, 10, 15, 20, 25, 30, 35 \mbox{ and so on to } 36 \times 5 = 180.$.

Dividing each of these by 5 gives

$1, 2, 3, 4, 5, 6, 7 \mbox{ and so on to } 36.$

Now you can see the problem, every time a multiple of $5 ^2 = 25$ divides 180 you are left with an extra 5. $25 \times 7 = 175$ so there are 7 multiples of 25 that divide 180! Hence 5 divides 180! at least $36 + 7 = 43$ times.

What about $5^{3} = 175?$ 175 divides 180 once so there is one more 5 that divides 181! and hence $5^{44}$ divides 181!.

Thus 44 is that largest positive integer for which $5^{k}$ divides 181!

What remains is to show that $(2^2)^{44}$ and $(3^2)^{44}$ divide 181!

Penny