How do you find three consecutive odd positive integers such that 3 times the sum of all three is 24 more than the product of the first and second integers?

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How do you find three consecutive odd positive integers such that 3 times the sum of all three is 24 more than the product of the first and second integers?

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Answer 1 · 2016-03-16T06:01:11

${1, 3, 5}$

Explanation:

Any odd number may be represented as $2k+1$ for some integer $k$ . Then, let's represent our odd integers as $2k+1, 2k+3, 2k+5$ where $k$ is a non-negative integer. From our initial condition, we have that

$(2k+1)(2k+3)+24 = 3((2k+1)+(2k+3)+(2k+5))$

Simplifying both sides, we get

$4k^2+8k+27 = 18k+27$

$=> 4k^2-10k = 0$

$=> 4k(k-5/2) = 0$

$=> k = 0$ or $k = 5/2$

But, as we specified that $k$ must be an integer, we know that $k!=5/2$ . Thus we have $k = 0$ and so our odd integers are ${1, 3, 5}$

Checking this, we find that $1*3+24 = 27 = 3(1+3+5)$

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How do you find three consecutive odd positive integers such that 3 times the sum of all three is 24 more than the product of the first and second integers?

1 Answer

Explanation:

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