Given that the tail of n! Has exactly 106 consecutive zeros, what is the maximum value of natural number n?

It should be the minimum
106 zeros need 106 5S
Then every time n is a multiple of 5, there are n / 5 5S
Every time n is a multiple of 25, there will be n / 25 more 5
Every time n is a multiple of 125, there will be n / 125 more 5
And so on, because only 106 are needed, n is less than 625
So | n / 5 | + | n / 25 | + | n / 125 | = 106
N = 427.6 multiple of 5

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