The number of zeros at the end of the product of all prime numbers between 1…

Concept: Trailing zeros in a product come from factors of 10 (= 2 x 5). The number of trailing zeros equals the number of matched pairs of the prime factors 2 and 5 present in the product's full factorization.

1. List the primes between 1 and 1111: 2, 3, 5, 7, 11, 13, 17, 19, 23, ... up to 1109.

2. Among all these primes, only the single prime 2 itself supplies a factor of 2 — every other prime is odd.

3. Among all these primes, only the single prime 5 itself supplies a factor of 5 — any other number ending in 5 (15, 25, 35, ...) is divisible by 5 and hence composite, not prime.

4. So the full prime factorization of the product contains the factor 2 exactly once and the factor 5 exactly once, giving exactly one factor of 10.

Cross-check: the number of trailing zeros equals the smaller of the power of 2 and the power of 5 in the factorization; both powers here are 1, so the trailing-zero count is 1 — consistent with the step-by-step count above.

Result: the product of all prime numbers between 1 and 1111 ends in exactly 1 zero.