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For how many integers n between 5 and 20...

For how many integers n between 5 and 20, inclusive, is the sum of 3n, 9n and 11n greater than 200?

A

4

B

8

C

12

D

16

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find how many integers \( n \) between 5 and 20, inclusive, satisfy the condition that the sum of \( 3n \), \( 9n \), and \( 11n \) is greater than 200. ### Step-by-Step Solution: 1. **Write the expression for the sum**: \[ S = 3n + 9n + 11n \] 2. **Combine like terms**: \[ S = (3 + 9 + 11)n = 23n \] 3. **Set up the inequality**: We want to find when this sum is greater than 200: \[ 23n > 200 \] 4. **Solve for \( n \)**: Divide both sides of the inequality by 23: \[ n > \frac{200}{23} \] Calculating \( \frac{200}{23} \): \[ \frac{200}{23} \approx 8.69 \] 5. **Determine the integer values of \( n \)**: Since \( n \) must be an integer, the smallest integer greater than 8.69 is 9. Therefore, \( n \) can take values from 9 to 20. 6. **List the integers from 9 to 20**: The integers are: 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. 7. **Count the integers**: To count the integers from 9 to 20: \[ 20 - 9 + 1 = 12 \] Thus, there are **12 integers** \( n \) between 5 and 20 for which the sum \( 3n + 9n + 11n \) is greater than 200. ### Final Answer: The answer is **12**. ---
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