# Twice the smallest of three consecutive odd integers is seven more than the largest, how do you find the integers?

May 6, 2016

Interpret the question and solve to find:

$11$, $13$, $15$

#### Explanation:

If the smallest of the three integers is $n$ then the others are $n + 2$ and $n + 4$ and we find:

$2 n = \left(n + 4\right) + 7 = n + 11$

Subtract $n$ from both ends to get:

$n = 11$

So the three integers are: $11$, $13$ and $15$.

May 6, 2016

The three consecutive odd integers are $11$, $13$ and $15$.

#### Explanation:

We are given 3 consecutive odd integers.

Let the first odd integer be $x$.

Then the next odd integer will be $x + 2$.

Since $x$ is odd, $x + 1$ will be even, and we want 3 odd integers that are consecutive.

The ${3}^{r d}$ integer will be $x + 2 + 2 = x + 4$

Now, We have our three integers, $x$, $x + 2$ and $x + 4$.

Clearly the smallest integer is $x$ and the largest is $x + 4$.

Given that: twice the smallest = 7 more than the largest.

$\implies 2 x = 7 + \left(x + 4\right)$

$\implies 2 x = x + 11$

$\implies x = 11$

Checking

Our 3 consecutive odd integers are $11$, $13$ and $15$.

Twice the smallest = $2 \times 11 = 22$

7 more than the largest = $7 + 15 = 22$