The sum of the squares of three consecutive odd numbers is 2531. Find the smallest of these numbers.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 27
Consecutive odd numbers differ by 2, so the three numbers can be written as \(n-2\), \(n\) and \(n+2\), with \(n\) the middle one. Written this way, the cross terms cancel when the squares are added.
\((n-2)^2+n^2+(n+2)^2=3n^2+8\)
\(3n^2+8=2531\)
\(3n^2=2523\)
\(n^2=841\), so \(n=29\)
The numbers are 27, 29 and 31, and the smallest is 27