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

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

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Answer 1 · 2017-01-12T08:51:31

$7, 9, 11$ $7, 9, 11$

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 $n$ $n$ , so the other two are $n + 2$ $n+2$ and $n + 4$ $n+4$ .

We are given:

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

which expands to:

$n^{2} + 2 n = n + 56$ $n^2+2n=n+56$

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

$0 = n^{2} + n - 56 = (n + 8) (n - 7)$ $0 = n^2+n-56 = (n+8)(n-7)$

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

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

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

Now consider negative integers.

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

We are given:

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

which expands to:

$n^{2} - 2 n = n + 48$ $n^2-2n=n+48$

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

$0 = n^{2} - 3 n - 48$ $0 = n^2-3n-48$

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

$0 = 4 (n^{2} - 3 n - 48)$ $0 = 4(n^2-3n-48)$

$0 = 4 n^{2} - 12 n - 192$ $color(white)(0) = 4n^2-12n-192$

$0 = {(2 n)}^{2} - 2 (2 n) (3) + 9 - 201$ $color(white)(0) = (2n)^2-2(2n)(3)+9-201$

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

$0 = {(2 n + 3)}^{2} - {(\sqrt{201})}^{2}$ $color(white)(0) = (2n+3)^2-(sqrt(201))^2$

$0 = ((2 n + 3) - \sqrt{201}) ((2 n + 3) + \sqrt{201})$ $color(white)(0) = ((2n+3)-sqrt(201))((2n+3)+sqrt(201))$

$0 = (2 n + 3 - \sqrt{201}) (2 n + 3 + \sqrt{201})$ $color(white)(0) = (2n+3-sqrt(201))(2n+3+sqrt(201))$

$0 = 4 (n + \frac{3}{2} - \frac{\sqrt{201}}{2}) (n + \frac{3}{2} + \frac{\sqrt{201}}{2})$ $color(white)(0) = 4(n+3/2-sqrt(201)/2)(n+3/2+sqrt(201)/2)$

So $n = - \frac{3}{2} \pm \frac{\sqrt{201}}{2}$ $n = -3/2+-sqrt(201)/2" "$ (not an integer)

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

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

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