How do you find three consecutive integers such that twice the smallest is 12 more than the largest?

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Answer 1 · 2016-01-22T00:24:19

Convert the conditions into an equation and solve to find that the three integers are: $14, 15, 16$

Suppose the three integers are $n$ , $n+1$ and $n+2$

We are given:

$2n = (n+2)+12$

Subtract $n$ from both sides to find:

$n = 2+12 = 14$

So the three integers are $14, 15, 16$ .