A favourite old chestnut of mathematicians, for today. It’s one of those puzzles that looks hard at first glance, but yields fairly easily to thought and logic.
Early on, all students of mathematics learn of the factorial function, which uses the exclamation mark (”!”). The factorial of any positive integer is the product of all the integers from 1 up to that one.
3! = 1 x 2 x 3 = 6
7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040
and so on.
Note that 7!, 5040, ends in a zero. We say it has one trailing zero.
Question: how many trailing zeros does 100! have?
Hint: What does a trailing zero mean? That it is a multiple of 10. You get at least one new trailing zero each time you multiply by 10 or a multiple of 10, for sure. So when you multiply by 10, 20, 30… 90 and 100, each will add a trailing zero. Thus 10 trailing zeros, right?
Except… are there other ways of adding a new trailing zero?
Consider how the trailing zero in 7!, 5040, appeared: when we multiplied by 5. (Check: 4! = 24. 5! = 120). That’s because we multiplied an even number (24) by 5 – which is the same as multiplying half that even number (12 in this case) by 10.
So actually, we get a new trailing zero every time we multiply by a multiple of 5. That’s 5, 10, 15… 95, 100. Thus 20 trailing zeros. Right?
Except… well, that’s enough of a hint. What’s the answer?
Scroll down for the answers.
20 from all those multiples of 5, yes. But 25 = 5 x 5, so when we multiply by 25 we are multiplying by 5 twice, giving us an extra trailing zero. The same applies to the multiples of 25: 50, 75 and 100. Thus 4 extra trailing zeros, for a total of… drum roll… 24.
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