# How do you find three consecutive odd integers such that four times the middle integer is twice the sum of the other two integers?

May 28, 2015

Nice question.

Any three consecutive odd integers have this property. (As do any three consecutive even integers.)

To solve this, let the consecutive odd integers be $n - 2$, $n$, and $n + 2$.

(We'll throw out any non-odd solutions.)

We want $4 \left(n\right) = 2 \left[\left(n - 2\right) + \left(n + 2\right)\right]$

But $\left(n - 2\right) + \left(n + 2\right) = 2 n$, so we want:

$4 \left(n\right) = 2 \left[\left(n - 2\right) + \left(n + 2\right)\right] = 2 \left[2 n\right] = 4 n$

We want: $4 n = 4 n$.

But for any number, $4 n = 4 n$, so any number (every number) is a solution to $4 \left(n\right) = 2 \left[\left(n - 2\right) + \left(n + 2\right)\right]$.

The question asks only for odd solutions, so:

Any (every) three consecutive odd integers will solve the problem.

We could have used $n , n + 2 , \text{and } n + 4$ as our 3 numbers.

Or we could make sure they are odd by making the numbers:

$2 k + 1$, $2 k + 3$, and $2 k + 5$ in whihc case, we solve:

$4 \left(2 k + 3\right) = 2 \left(2 k + 1 + 2 k + 5\right)$ and we get:

$8 k + 12 = 2 \left(4 k + 6\right) = 8 k + 12$

Again, every number $k$ gives a solution.