What is the number of zeros at the end of the product of the numbers from $1$ to $100$?

### Concept & Logic The number of trailing zeros in a factorial or continuous product is determined by the number of $(2 \times 5)$ pairs. Because multiples of $2$ are far more frequent than multiples of $5$, the bottleneck is the highest power of $5$. ### Step-by-Step Solution **Given:** The product is $1 \times 2 \times 3 \times \dots \times 100$, which is $100!$. **Calculation / Deduction:** * We need to find the highest power of $5$ in $100!$. * We use Legendre's formula (successive division) to count the multiples of $5$, $25$, $125$, etc. $$ \left\lfloor \frac{100}{5} \right\rfloor + \left\lfloor \frac{100}{5^2} \right\rfloor $$ * Calculate the terms: $$ \frac{100}{5} = 20 $$ $$ \frac{100}{25} = 4 $$ * Add the results: $20 + 4 = 24$. ### Exam Strategy & Shortcut Use the successive division method mentally. Divide $100$ by $5$ to get $20$. Then divide that quotient ($20$) by $5$ to get $4$. Since $4$ is less than $5$, stop. Add the quotients: $20 + 4 = 24$. ### Common Pitfall A frequent error is stopping after the first division (yielding $20$) and forgetting that multiples of $25$ (like $25, 50, 75, 100$) contain an extra $5$ that must be counted. Always continue dividing the quotient until it is less than $5$. ### Final Answer Therefore, the correct answer is 24.