First 100 multiples of 10 i.e. $10, 20, 30, ........., 1000$ are multiplied together. The number of zeros at the end of the product will be

### Concept & Logic When dealing with a sequence of multiples, you must factor out the common base from *every single term* in the product. After factoring, the problem usually reduces to evaluating a standard power of $10$ combined with a standard factorial sequence. ### Step-by-Step Solution * **Given:** The product sequence is $10 \times 20 \times 30 \times \dots \times 1000$. There are $100$ terms in this sequence ($1000 \div 10 = 100$). * **Calculation / Deduction:** Rewrite each term in the sequence as a multiple of $10$: $$ (10 \times 1) \times (10 \times 2) \times (10 \times 3) \times \dots \times (10 \times 100) $$ Factor out the $10$ from each of the $100$ terms. Because these terms are multiplied (not added), the $10$ is raised to the power of $100$: $$ 10^{100} \times (1 \times 2 \times 3 \times \dots \times 100) $$ $$ = 10^{100} \times 100! $$ Now, calculate the number of trailing zeros from both components: 1. **Zeros from $10^{100}$**: A base of $10$ raised to the power of $100$ inherently provides exactly **$100$** trailing zeros. 2. **Zeros from $100!$**: Use successive division by $5$ to find the trailing zeros in $100!$: $$ \lfloor 100 / 5 \rfloor = 20 $$ $$ \lfloor 20 / 5 \rfloor = 4 $$ Total zeros in $100! = 20 + 4 =$ **$24$** zeros. Add the zeros from both components together to find the total for the entire expression: $$ \text{Total trailing zeros} = 100 + 24 = 124 $$ ### Exam Strategy & Shortcut **Decomposition Visualization:** Instantly visualize the series as two layers: the visible zeros and the hidden zeros. 1. There are $100$ numbers, and each clearly ends in at least one zero. That's $100$ guaranteed zeros immediately. 2. Strip away those $100$ zeros, and you are left with $1 \times 2 \times \dots \times 100$, which is $100!$. 3. Calculate zeros for $100!$ mentally: $100 \div 5 = 20$; $20 \div 5 = 4$. So, $24$. 4. Total $= 100 + 24 = 124$. ### Common Pitfall The most dangerous pitfall is factoring out a single $10$ as if the expression was an addition series (e.g., mistakenly writing $10 \times (1 \times 2 \dots \times 100)$). Because it is a multiplication series, you must factor out $10$ for *each* term, creating $10^{100}$. ### Final Answer **Therefore, the correct answer is 124.**