Find the number of zeros at the end of the result $3 \times 6 \times 9 \times 12 \times 15 \times \dots \times 99 \times 102$.

### Concept & Formula Similar to counting zeros in factorials, when dealing with an arithmetic progression of multiples, extract the common factor to reveal a standard factorial. Count the pairs of $2$s and $5$s. ### Step-by-Step Solution **Given:** Product is $3 \times 6 \times 9 \times 12 \times 15 \times \dots \times 102$. **Calculation:** * Note that each term is a multiple of $3$: $(3 \times 1) \times (3 \times 2) \times \dots \times (3 \times 34)$. * Factor out the $3$ from all $34$ terms: $$3^{34} \times (1 \times 2 \times 3 \times \dots \times 34)$$ * This simplifies to $3^{34} \times 34!$. * The number of zeros is determined by the number of $5$s in $34!$, as $2$s are abundant and the $3^{34}$ term contributes no $2$s or $5$s. * Count the highest power of $5$ in $34!$: $$\lfloor \frac{34}{5} \rfloor + \lfloor \frac{34}{25} \rfloor = 6 + 1 = 7$$ ### Exam Strategy & Shortcut Identify the multiple ($3$), determine the number of terms ($102 / 3 = 34$), and then find the highest power of $5$ in $34!$ using successive division ($34 / 5 = 6$, $6 / 5 = 1$, sum $= 7$). This can be done mentally in seconds. ### Common Pitfall Forgetting to check the power of $2$ can be risky. While in $34!$ there are always more $2$s than $5$s, if the sequence was purely multiples of $5$ (e.g., $5, 10, 15\dots$), you would need to count the $2$s instead. Always confirm which limiting factor dictates the pairs. ### Final Answer **Therefore, the correct answer is 7.**