Question

How do you find three consecutive odd integers such that the product of the two smaller exceeds the largest by 52?

Answer 1

`7, 9, 11`

Explanation:

Note the question speaks of "smaller" and "largest" numbers rather than "lesser" and "greatest", so it is easier to split positive and negative cases...

`color(white)()`
Case - positive integer solutions

Considering positive integers first, let the smallest be denoted by `n`, so the other two are `n+2` and `n+4`.

We are given:

`n(n+2) = (n+4)+52`

which expands to:

`n^2+2n=n+56`

Subtract `n+56` from both sides to get:

`0 = n^2+n-56 = (n+8)(n-7)`

Which gives us `n=7` or `n=-8`.

Since we specified positive integers, the three numbers are `7, 9, 11`

`color(white)()`
Case - negative integer solutions

Now consider negative integers.

If the smallest negative integer is `n`, then the other two are `n-2` and `n-4`.

We are given:

`n(n-2)=(n-4)+52`

which expands to:

`n^2-2n=n+48`

Subtract `n+48` from both sides to get:

`0 = n^2-3n-48`

Multiply by `4` so we can complete the square with integers:

`0 = 4(n^2-3n-48)`

`color(white)(0) = 4n^2-12n-192`

`color(white)(0) = (2n)^2-2(2n)(3)+9-201`

`color(white)(0) = (2n+3)^2-201" "` (I could stop here, but...)

`color(white)(0) = (2n+3)^2-(sqrt(201))^2`

`color(white)(0) = ((2n+3)-sqrt(201))((2n+3)+sqrt(201))`

`color(white)(0) = (2n+3-sqrt(201))(2n+3+sqrt(201))`

`color(white)(0) = 4(n+3/2-sqrt(201)/2)(n+3/2+sqrt(201)/2)`

So `n = -3/2+-sqrt(201)/2" "` (not an integer)

There are no (negative) integer solutions since `201` is not a perfect square.