The numbers $1, 3, 5, 7, ........., 99$ and $128$ are multiplied together. The number of zeros at the end of the product must be

### Concept & Logic Every trailing zero in a product requires exactly one pair of the prime factors $2$ and $5$. In this problem, the product consists of two distinct parts: 1. A continuous series of purely **odd numbers** ($1 \times 3 \times 5 \dots \times 99$). 2. A single standalone even number appended at the end ($128$). Odd numbers contain zero $2$'s. Therefore, the only $2$'s available in the entire product are locked inside the number $128$. Because $2$'s will be extremely scarce compared to the abundance of $5$'s in the odd series, $2$ becomes the limiting factor. ### Step-by-Step Solution * **Given:** Product: $(1 \times 3 \times 5 \times 7 \times \dots \times 99) \times 128$ * **Calculation / Deduction:** **Step 1: Count the available 2's.** The entire odd series contributes zero $2$'s. The number $128$ is a known power of $2$: $$ 128 = 2^7 $$ Therefore, the entire expression contains exactly **$7$** twos. **Step 2: Count the available 5's (Verification).** To be thorough, we should ensure there are at least seven $5$'s in the odd series. The series contains odd multiples of $5$: $5, 15, 25, 35, 45, 55, 65, 75, 85, 95$. Even just looking at the first few: $5$ (one), $15$ (one), $25$ (two), $35$ (one), $45$ (one)... we clearly have more than seven $5$'s available. **Step 3: Determine the limiting factor.** We have an abundance of $5$'s (more than $12$), but exactly $7$ twos. Because we can only form pairs up to the limit of the scarcest resource, we can only form exactly **$7$** pairs of $(2 \times 5)$. ### Exam Strategy & Shortcut **Identify the Restrictor:** Rapid analysis of the question reveals the trick immediately. The huge odd series is a distractor meant to make you count $5$'s needlessly. Recognize that odd numbers have no $2$'s. Look at the only even number: $128$. Recognize that $128$ is $2^7$. The maximum number of $10$'s you can make is locked at $7$. Done. ### Common Pitfall The most common trap is ignoring the specific nature of the series, automatically calculating the $5$'s in numbers up to $99$ using Legendre's successive division (which is also wrong here since it's missing even numbers), and completely missing that the total zeros are restricted entirely by the single even number appended at the end. ### Final Answer **Therefore, the correct answer is 7.**