Question

The sum of three consecutive even integers is 138. Find the integers.

Answer 1

Let \(n\) be the smallest among the three.
Then \(n+(n+1)+(n+2)=138\)
\(3n+3=138\)
\(3n=135\)
\(n=45\)
So the integers are \(45,46,47\).

Answer 2

@watchman , 45 and 47 are not even.

Answer 3

ha ha sorry ... so forget my answer :)

Answer 4

\(n+(n+2)+(n+4)=138\)
\(3n+6=138\)
\(3n=132\)
\(n=44\)
So \(44,46,48\)