How do you find three consecutive odd integers whose sum is 13 more than twice the largest of the three?

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Answer 1 · 2016-01-24T01:02:48

Interpret the description in terms of an unknown smallest integer $n$ and solve to find integers:

$15, 17, 19$

Suppose the three odd integers are $n$ , $n+2$ and $n+4$

Their sum is:

$n + (n+2) + (n+4) = 3n+6$

$13$ more than twice the largest of the three is:

$2(n+4)+13 = 2n+21$

From what we are told these two are equal:

$3n+6 = 2n+21$

Subtract $2n+6$ from both sides to get:

$n = 15$

So the three integers are:

$15, 17, 19$