Question

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

Answer 1

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(n) = 2[(n-2)+(n+2)]`

But `(n-2)+(n+2) = 2n`, so we want:

`4(n) = 2[(n-2)+(n+2)] = 2[2n] = 4n`

We want: `4n=4n`.

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

The question asks only for odd solutions, so:

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

Additional Notes

We could have used `n, n+2, "and " n+4` as our 3 numbers.

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

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

`4(2k+3) = 2(2k+1+2k+5)` and we get:

`8k+12 = 2(4k+6) = 8k+12`

Again, every number `k` gives a solution.