Question

A particular even integer, `n`, and the next even integer after it is less than or equal to the average of that second even integer and the one after *it*, plus `7`. What is the largest integer for which this is true?

Answer 1

`n<=8`

Explanation:

We can set this up as an inequality and solve it.

If we call the first even integer `n` then the next consecutive even integer will be `n+2`. The following even integer will be `n+4`, the next `n+6` and so on.

We can set this up as an inequality:

`n+(n+2)<=((n+2)+(n+4))/2+7`

This exactly translates what the question asks. Now it's just algebra!

Collect like terms on each side:

`2n+2<=(2n+6)/2+7`

Multiply through by 2:

`4n+4<=2n+6+14`

`4n+4<=2n+20`

Collect terms in `n` on the left and numbers on the right:

`2n<=16`

Divide both sides by 2:

`n<=8`

If you wanted to, you could try the even numbers less than or equal to 8 (i.e. 2, 4, 6, 8) in the inequality to make sure it's true.