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How do you find three consecutive odd integers such that the sum of all three is 36 less than the product of the smaller two?

1 Answer

Dec 11, 2016

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

Explanation:

Let the numbers be $x$ $x$ , $x + 2$ $x + 2$ and $x + 4$ $x + 4$

$x + (x + 2) + (x + 4) = x (x + 2) - 36$ $x + (x + 2) + (x + 4) = x(x + 2) - 36$

$x + x + 2 + x + 4 = x^{2} + 2 x - 36$ $x + x + 2 + x + 4 = x^2 + 2x - 36$

$3 x + 6 = x^{2} + 2 x - 36$ $3x + 6 = x^2 + 2x- 36$

$0 = x^{2} - x - 42$ $0 = x^2 - x - 42$

$0 = (x - 7) (x + 6)$ $0 = (x - 7)(x + 6)$

$x = 7 and - 6$ $x = 7 and -6$

Since the numbers have to be odd, the numbers are $7$ $7$ , $9$ $9$ and $11$ $11$ .

Hopefully this helps!

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