Trailing Zeros

Problem

How many trailing zeros are there in 100! (factorial of 100)?

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Hints

Hint 1

Think about prime factorization of factorial numbers.

Hint 2

Which number is rarer in forming 10s, the factor 2 or the factor 5?

Hint 3

Be careful to count multiples of 25, 50, 75, and 100 because they contribute extra factors of 5 beyond the first.

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Solution

Show Solution

24

Show Explanation

Trailing zeros in a factorial come from factors of 10. Every 10 is made by multiplying 2 × 5. In a factorial like 100!, there are always plenty of 2's from even numbers, so the limiting factor is how many 5's you can find.

• First, count how many multiples of 5 are between 1 and 100: there are $\lfloor 100/5 \rfloor = 20$ such numbers (5, 10, 15, …, 100).
• But some numbers contribute more than one 5 in their prime factorization. Specifically, multiples of 25 (like 25, 50, 75, 100) contribute an extra factor of 5, because $25 = 5^2$. That adds another $\lfloor 100/25 \rfloor = 4$ factors of 5.

So the total number of 5's is $20 + 4 = 24$, and each pairs with a 2 to make a trailing zero. Therefore 100! ends with 24 trailing zeros.

Follow-up Question

Follow-up

What if it is $n!$ ? Could you solve it?