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:

For consecutive integers, set up the equation with the middle integer as a variable to simplify calculations.

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: $3n^2 + 6n - 105 = 0$, or $n^2 + 2n - 35 = 0$.
- Step 4: Solve quadratic: Discriminant = $4 + 140 = 144$, $n = \dfrac{-2 \pm 12}{2} = 5$ or $-7$.
- Step 5: Test $n = 5$: $5^2 + 6^2 + 7^2 = 25 + 36 + 49 = 110$. Smallest is 5. Test $n = -7$: $(-7)^2 + (-6)^2 + (-5)^2 = 49 + 36 + 25 = 110$, smallest is $-7$. Since options are positive, choose 5.
- Step 6: Check options: Option (a) is 5, which matches.