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: .Find all positive integers such that 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 ?
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 ends in 72 zeros,
. . but ends in 74 zeros.
. .
And the same is true for
. . But ends in 75 zeros.
Therefore: .