What is the number of zeros at the end of the product $5^5 \times 10^{10} \times 15^{15} \times \dots \times 125^{125}$?

### Concept & Strategy Trailing zeros require pairs of $(2 \times 5)$. In standard factorials, $2$ is abundant, so we count $5$s. However, in this specific product, every term is a multiple of $5$ raised to large powers, meaning $5$s are highly abundant. The bottleneck here is actually the highest power of $2$. ### Step-by-Step Solution **Given:** Product $N = 5^5 \times 10^{10} \times \dots \times 125^{125}$. **Calculation / Deduction:** * We must find the highest power of $2$ in this product. Only terms that are even will contribute a factor of $2$. These are $10^{10}, 20^{20}, 30^{30}, \dots, 120^{120}$. * The power of $2$ contributed is the sum of their exponents, but we must also account for multiples of $4$ ($2^2$), $8$ ($2^3$), and $16$ ($2^4$). * **Multiples of 2:** $10 + 20 + 30 + \dots + 120 = 10(1 + 2 + \dots + 12) = 10 \times \frac{12 \times 13}{2} = 780$. * **Multiples of 4:** Provide an additional factor. These are $20^{20}, 40^{40}, \dots, 120^{120}$. Sum: $20 + 40 + \dots + 120 = 20(1 + 2 + \dots + 6) = 20 \times \frac{6 \times 7}{2} = 420$. * **Multiples of 8:** Provide a third factor. These are $40^{40}, 80^{80}, 120^{120}$. Sum: $40 + 80 + 120 = 240$. * **Multiples of 16:** Provide a fourth factor. The only term is $80^{80}$. Sum: $80$. * Total power of $2 = 780 + 420 + 240 + 80 = 1520$. ### Exam Strategy & Shortcut Whenever a sequence is heavily weighted with multiples of $5$, instantly switch your focus to finding the total power of $2$. Group the exponents using Arithmetic Progression (AP) sums to calculate the totals rapidly instead of adding them manually. ### Common Pitfall Assuming that the power of $5$ dictates the number of trailing zeros in *every* problem. If you try to find the power of $5$ here, you will get a massive number, leading you to an incorrect assumption and a wrong answer. Always identify the limiting prime factor first. ### Final Answer Therefore, the correct answer is 1520.