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
Hello, silentbob!
This takes a bit of brainwork . . .
We know that: .$\displaystyle n! \;=\;1\cdot2\cdot3\cdot4 \cdots n$Find all positive integers $\displaystyle n$ such that $\displaystyle n!$ ends with exactly 74 zeroes.
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 $\displaystyle n!$ ?
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 $\displaystyle 299!$ ends in 72 zeros,
. . but $\displaystyle 300!$ ends in 74 zeros.
. . $\displaystyle \begin{array}{ccc}\left[\dfrac{300}{5}\right] &=& 60 \\ \\[-3mm]
\left[\dfrac{300}{25}\right] &=& 12 \\ \\[-3mm]
\left[\dfrac{300}{125}\right] &=& 2 \\ \\[-3mm] \hline \\[-3mm]
\text{Total:} & & 74 \end{array}$
And the same is true for $\displaystyle 301!,\:302!,\:303!,\:304!$
. . But $\displaystyle 305!$ ends in 75 zeros.
Therefore: .$\displaystyle n \;=\;300,\:301,\:302,\:304$