The numbers $1, 3, 5, ........, 25$ are multiplied together. The number of zeros at the right end of the product is

### Concept & Logic A trailing zero at the right end of any product is created exclusively by multiplying the prime factors $2$ and $5$ together ($2 \times 5 = 10$). The number of trailing zeros is determined by the number of pairs of $(2 \times 5)$ that can be formed from the prime factorization of all numbers in the series. ### Step-by-Step Solution * **Given:** A series of numbers multiplied together: $1 \times 3 \times 5 \times 7 \times .... \times 25$. * **Calculation / Deduction:** Observe the nature of the numbers in the series. $1, 3, 5, 7, 9, 11...$ all the way to $25$. This is a continuous sequence of **odd numbers**. By definition, an odd number does not have $2$ as a prime factor. While the series contains plenty of $5$'s (in $5$, $15$, and $25$), it contains absolutely zero $2$'s. Without any $2$'s to pair with the $5$'s, it is mathematically impossible to form a $10$. Therefore, the final product will not have any trailing zeros. ### Exam Strategy & Shortcut **Parity Rule:** The product of any number of odd integers is ALWAYS an odd integer. An odd integer can never end in the digit $0$ (since ending in $0$ inherently makes a number even and divisible by $10$). Therefore, the number of zeros at the end is $0$. This deduction should take less than 5 seconds. ### Common Pitfall Students who have memorized the standard "Trailing Zeros" formula often operate on autopilot. They will scan the series, correctly identify the multiples of $5$, count them up, and mistakenly conclude there are several trailing zeros, completely ignoring the fact that the crucial even numbers are missing from the given sequence. ### Final Answer **Therefore, the correct answer is 0.**