https://file.aiquickdraw.com/mathbot/file/17350678954763exfsgwc.jpg

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The problem asks:

If \( n = 2^8 \cdot 3^4 \cdot 5^3 \cdot 7 \), then the number of **consecutive zeros at the end of \( n \)** is to be determined.

### Solution:

The number of consecutive zeros at the end of a number is determined by the number of **factors of 10** in the number. A factor of 10 is formed by multiplying one factor of 2 and one factor of 5.

#### Step 1: Count factors of 2 and 5
- From \( n = 2^8 \cdot 3^4 \cdot 5^3 \cdot 7 \):
  - The power of 2 is 8.
  - The power of 5 is 3.

#### Step 2: Pair factors of 2 and 5
Each pair of one 2 and one 5 contributes to one factor of 10. Therefore, the number of factors of 10 is the **minimum of the powers of 2 and 5**:
\[
\text{Number of factors of 10} = \min(8, 3) = 3
\]

#### Step 3: Conclusion
The number \( n \) has **3 consecutive zeros** at the end.

### Final Answer:
**(B) 3**

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Would you like further clarification? Here are 5 related questions to explore:

1. How do powers of prime factors affect the trailing zeros in a number?
2. What would the result be if \( n = 2^{10} \cdot 5^4 \)?
3. How can we calculate the total number of factors of a number \( n \)?
4. What is the role of other prime factors (e.g., 3 and 7) in determining trailing zeros?
5. Can trailing zeros ever result from numbers without the factor 5?

### Tip:
Always focus on the pairing of 2s and 5s to determine trailing zeros, as 10 is the key factor here.

If n = 2^8 * 3^4 * 5^3 * 7, then the number of consecutive zeros at the end of the number.

Learn how to determine the number of trailing zeros in the number n = 2^8 * 3^4 * 5^3 * 7. Explore key concepts in number theory and factors, with a detailed step-by-step explanation. Suitable for students in Grades 9-12.

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