The sum of four consecutive integers is 1290. What is the greatest of them?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 324
Four consecutive integers can be written as \(n\), \(n+1\), \(n+2\) and \(n+3\); their sum gives \(n\).
\(n+(n+1)+(n+2)+(n+3)=4n+6\)
\(4n+6=1290\), so \(4n=1284\) and \(n=321\)
The integers are 321, 322, 323 and 324
The greatest of them is 324.