Question

Let $S$ be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all elements of $S$. With how many consecutive zeros will the product end?

• A. 1
• B. 4
• C. 5
• D. 10

Hint:

In trailing zero problems, the limiting factor is the number of 5s in the prime factorisation.

Answer 1

The Correct Option is A

Trailing zeros in a number come from factors of $10 = 2 \times 5$. The product of all primes below 100 will contain: - Exactly one factor 2 (since 2 is prime). - Exactly one factor 5 (since 5 is prime). Since each 10 requires one 2 and one 5, and we have only one 5, we can form only one factor of 10. \[ \boxed{\text{Number of trailing zeros} = 1} \]