The numbers $2, 4, 6, 8, \dots, 98, 100$ are multiplied together. The number of zeros at the end of the product must be

### Concept & Formula The number of zeros at the end of a product is determined by the number of trailing tens, which equals the number of pairs of prime factors $2$ and $5$. ### Step-by-Step Solution **Given:** The sequence is the product of even numbers: $2 \times 4 \times 6 \times \dots \times 100$. **Calculation:** * Factor out a $2$ from each of the $50$ terms in the product: $$2^{50} \times (1 \times 2 \times 3 \times \dots \times 50)$$ * This simplifies to $2^{50} \times 50!$. * To find the number of trailing zeros in $50!$, we only need to count the highest power of $5$, since the power of $2$ (which includes $2^{50}$) is vastly greater. * Use Legendre's formula to count the $5$s in $50!$: $$\lfloor \frac{50}{5} \rfloor + \lfloor \frac{50}{25} \rfloor = 10 + 2 = 12$$ * There are $12$ fives and more than enough twos to pair them with. Thus, there are $12$ pairs of $(2, 5)$. ### Exam Strategy & Shortcut Whenever you have a continuous series of multiples (like $2, 4, 6\dots$ or $3, 6, 9\dots$), factor out the common multiple raised to the power of the number of terms. Then simply use the standard zero-counting rule on the resulting factorial part. ### Common Pitfall A frequent error is trying to apply the zero-counting rule directly to $100$ as if it were $100!$, which would incorrectly yield $24$ zeros. Always factor out the sequence properly first. ### Final Answer **Therefore, the correct answer is 12.**