Question

Which of the following sets of three positive consecutive multiples of 3 satisfy the condition that the largest number times itself is at least 81 more than the product of the smaller two?

A) 6, 9, 12
B) 9, 12, 15
C) 12, 15, 18
D) 15, 18, 21
E) 18, 21, 24

Answer 1

The correct set of three positive consecutive multiples of 3 that satisfies the condition is A) 6, 9, 12. By setting up an inequality and solving it, we find that the sequence starting with 6 fits the required condition.

Step-by-step explanation:

The student's question asks which set of three positive consecutive multiples of 3 satisfies the condition that the largest number squared is at least 81 more than the product of the smaller two. To find the answer, we can set up an inequality using algebraic expressions.

Let the smallest multiple be 3n, where n is a positive integer. Then the three numbers are 3n, 3(n+1), and 3(n+2). The condition states that:

(3(n+2))^2 ≥ 3n * 3(n+1) + 81

Simplifying both sides, we get:

9(n+2)^2 ≥ 9n(n+1) + 81

n^2 + 4n + 4 ≥ n^2 + n + 9

3n ≥ 5

Since n must be a positive integer, the smallest n that satisfies this inequality is 2. Therefore, the multiples are 6, 9, and 12. We can verify this by calculating:

• (3(2+2))^2 = 144
• 3(2) * 3(2+1) = 6 * 9 = 54
• 144 is indeed at least 81 more than 54.

This means the correct answer is A) 6, 9, 12.