Question

The sum of the squares of three consecutive integers is 110. What is the smallest integer?

• A. 5
• B. 6
• C. 7
• D. 8

Hint:

Form a quadratic equation for consecutive integers and solve for the smallest.

Answer 1

The Correct Option is A

- Step 1: Let the integers be $n, n+1, n+2$. Given: $n^2 + (n+1)^2 + (n+2)^2 = 110$.
- Step 2: Expand: $n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4) = 3n^2 + 6n + 5 = 110$.
- Step 3: Simplify: $3n^2 + 6n - 105 = 0 \implies n^2 + 2n - 35 = 0$.
- Step 4: Solve: $n = \frac{-2 \pm \sqrt{4 + 140}}{2} = \frac{-2 \pm 12}{2}$, so $n = 5$ or $n = -7$.
- Step 5: If $n = 5$, numbers are 5, 6, 7. Sum = $5^2 + 6^2 + 7^2 = 25 + 36 + 49 = 110$. If $n = -7$, numbers are -7, -6, -5, sum = $49 + 36 + 25 = 110$. Smallest is -7 or 5.
- Step 6: Option (1) is 5, correct.