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

### Concept & Logic The number of zeros at the end of a product strictly depends on the number of pairs of the prime numbers $2$ and $5$ present in its prime factorization. ### Step-by-Step Solution **Given:** $S$ is the set of prime numbers $\ge 2$ and $< 100$. The product is $2 \times 3 \times 5 \times 7 \times 11 \dots \times 97$. **Deduction:** * The prime numbers up to $100$ include $2, 3, 5, 7, 11, \dots$ * A trailing zero is formed entirely by multiplying $2 \times 5 = 10$. * Since we are multiplying a set containing *only* prime numbers exactly once, the prime factor $2$ appears exactly once. * The prime factor $5$ also appears exactly once. * No other combinations of primes will yield a factor of $10$. * Since there is only one $2$ and one $5$, we can only form exactly one pair of $(2, 5)$. ### Exam Strategy & Shortcut This is a pure logic question. Primes don't have factors. In any sequence of consecutive primes starting from $2$, there will always be exactly one $2$ and one $5$. Therefore, the product of any set of consecutive primes that includes both $2$ and $5$ will always end in exactly $1$ zero, no matter how many other primes you multiply. ### Common Pitfall Students often overthink the size of the primes up to $100$ and assume a large product will naturally have multiple zeros, forgetting that primes (other than 2 and 5) never contribute to trailing zeros. ### Final Answer **Therefore, the correct answer is 1.**