The number of zeros at the end of the product $5 \times 10 \times 15 \times 20 \times 25 \times 30 \times 35 \times 40 \times 45 \times 50$ is

### Concept & Logic Trailing zeros are formed by pairs of $(2 \times 5)$. In standard continuous series (like factorials), $5$'s are always rarer than $2$'s, so we usually only count the $5$'s. However, **in a series composed entirely of multiples of 5**, the environment is flooded with $5$'s. In this specific scenario, the prime factor **$2$ becomes the limiting factor**. The number of trailing zeros will be dictated strictly by the number of $2$'s available to pair with the abundant $5$'s. ### Step-by-Step Solution * **Given:** Product: $5 \times 10 \times 15 \times 20 \times \dots \times 50$ This is a sequence of the first $10$ multiples of $5$. * **Calculation / Deduction:** Rewrite the sequence by factoring out $5$ from each term: $$ (5 \times 1) \times (5 \times 2) \times (5 \times 3) \times \dots \times (5 \times 10) $$ $$ = 5^{10} \times (1 \times 2 \times 3 \times \dots \times 10) $$ $$ = 5^{10} \times 10! $$ Now, let's analyze the prime factors $2$ and $5$ in this expression: **Count the 5's:** There are $10$ fives from the $5^{10}$ term. In $10!$, there are $\lfloor 10/5 \rfloor = 2$ fives. Total $5$'s available $= 10 + 2 = 12$. **Count the 2's:** The $5^{10}$ term contains zero $2$'s. All the $2$'s must come entirely from the $10!$ component. Use successive division to count the $2$'s in $10!$: $$ \lfloor 10 / 2 \rfloor = 5 $$ $$ \lfloor 5 / 2 \rfloor = 2 $$ $$ \lfloor 2 / 2 \rfloor = 1 $$ Total $2$'s available $= 5 + 2 + 1 =$ **$8$**. Since a trailing zero requires a *pair* of $(2 \times 5)$, and we only have eight $2$'s available to pair with our twelve $5$'s, we can only form $8$ pairs. ### Exam Strategy & Shortcut **Identify the Bottleneck:** The moment you see a series jumping by multiples of $5$, your brain should flag: "2 is the bottleneck!" Don't even waste time counting the $5$'s. Just factor out the $5$ to find the underlying factorial: $5 \times \dots \times 50 = 5^{10} \times 10!$. Then, strictly count the number of $2$'s in $10!$ via $10 \div 2 = 5$, $5 \div 2 = 2$, $2 \div 2 = 1$. Sum $= 8$. You are done in 15 seconds. ### Common Pitfall The classic trap is calculating the number of $5$'s ($12$) and selecting an option based on that, purely out of habit from solving standard factorial problems. Always verify which prime factor is the actual limiting constraint in a custom series. ### Final Answer **Therefore, the correct answer is 8.**