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The symbol $50!$ represents the product of all the whole numbers from $1$ to $50$ inclusive; that is, $50!=1 \times 2 \times 3 \times \dots \times 49 \times 50$. If I were to calculate the actual value, how many zeros would the answer have at the end?
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!