Can someone help me solve this problem? Thank you!

LOL @vidishas99 We both deleted at the same time!

Here I am waiting for your solutions and looking it up on wolframalpha at the same time

Adding again @HoldMyBeer

The number of zeroes indicate the number of times you can factor out 10 from a number

10 = 2 * 5

25! has a lot of 2s in it:

2,4,6,8,10 alone give 8 2s

So 5 is the limiting factor, the number of 5s in 25!:

1 (5) + 1(10)+ 1 (15) + 1(20) + 2(25) = 6

Number of 10s = 6

So number of zeroes at the end of 25! = 6

Oh maaaan, I did figure out 5 but when I saw it had 6, that was such a bummer. Nicely explained!

Thanks a lot!

If you have 125!..then the number of 0s will beâ€¦ 25 (All multiples of 5 till 125) +5 (Five 5â€™s for all 25 multiples)+1 (one five extra for 125)

The general formula to find the number of factors of a prime number p in the factorial of a number n,

number of factors of p = [\,\frac{n}{p}]\, + [\,\frac{n}{p^2}]\, + [\,\frac{n}{p^3}]\,+ ...

where [x] refers to the â€śstep of xâ€ť or â€śgreatest integer less than or equal to xâ€ť

According to gregâ€™s previous solutions, I was supposed to solve this like 25!/ y = integer. However, as there is no value for y. I mean you said that zeroâ€™s are indicating 10 but I donâ€™t know why I should choose 5 as the limiting factor.

Because 25! has plenty of 2s

2,4,6,8,10 alone give me 8 2s

Compare that to 5s

There are a total of 6 5s

So you can multiply each of these 5s to 2 and get a 6 10s

Which would give you 6 0s at the end