How many zeroes are there at the end of the following product?
1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60

Correct Answer: (A) 10

The number of trailing zeroes is determined by $\min(v_2, v_5)$. Counting the factors of 2 in the even multiples: $10(2^1), 20(2^2), 30(2^1), 40(2^3), 50(2^1), 60(2^2)$, giving $1 + 2 + 1 + 3 + 1 + 2 = 10$ factors of 2. Although there are 14 factors of 5, the total number of complete pairs of $(2$$\times$ 5) is limited to 10, producing exactly 10 trailing zeroes.