Find the number of factors of 5 in natural number n

factors.java

public static int factorsOfFiveIn(int n) {
    int factorsOfFive = 0;
    while (n > 1) {
        if (n % 5 == 0) factorsOfFive++;
        n /= 5;
    }
    return factorsOfFive;
}

@EffectiveLabs Not quite. 25 has the factor 5 twice, so 25! adds 2 zeroes to the end compared to 24!.

I tried to write what you describe with Vavr (Java FP library) and somehow I got it wrong, which led me in the direction I went in the end. I had tried to map n to n%5==0 and then count all the true values. I still don't know how I got that wrong.

@jonsweimarck Thank you for this! I really like the version with the anonymous functions. I didn't know about iterate(), and Vavr's version also has it. This gives me a direction in which to refactor.

Using Vavr's implementations: divide n by 5 as long as n is a multiple of 5, then count the number of times you successfully divided n by 5.

    public static int factorsOfFiveIn(int n) {
        return Stream.iterate(n, divideBy(5))
                .map(isAMultipleOf(5))
                .takeWhile(isTrue())
                .length();
    }

As a post script, I found an alternative implementation of the larger exercise, which makes factorsOfFiveIn() obsolete. :) I'll publish the whole thing somewhere quite soon.