1. ## quick question

the problem is as follows, i have NO idea how to approach it or do it. any help is appreciated.

find all positive integers n such that n! ends with exactly 74 zeroes in decimal notation

2. Hello, silentbob!

This takes a bit of brainwork . . .

Find all positive integers $n$ such that $n!$ ends with exactly 74 zeroes.
We know that: . $n! \;=\;1\cdot2\cdot3\cdot4 \cdots n$

A final zero is created by a factor of 5 (and an even number).
The question becomes: how many factors of 5 are contained in $n!$ ?

We know that every 5th number is a multiple of 5.

Also that every 25th number is a multiple of 5² = 25,
. . each of which contributes another 5.

And that every 125th number is a multiple of 5³ = 125,
. . each of which contributes yet another 5.

By trial-and error, we find that $299!$ ends in 72 zeros,

. . but $300!$ ends in 74 zeros.

. . $\begin{array}{ccc}\left[\dfrac{300}{5}\right] &=& 60 \\ \\[-3mm]

\left[\dfrac{300}{25}\right] &=& 12 \\ \\[-3mm]

\left[\dfrac{300}{125}\right] &=& 2 \\ \\[-3mm] \hline \\[-3mm]
\text{Total:} & & 74 \end{array}$

And the same is true for $301!,\:302!,\:303!,\:304!$
. . But $305!$ ends in 75 zeros.

Therefore: . $n \;=\;300,\:301,\:302,\:304$