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

2 Answers
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:

2n = (n+4)+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^(rd) 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.

=>2x = 7 + (x + 4)

=>2x = x + 11

=>x = 11

Checking

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

Twice the smallest = 2xx11=22

7 more than the largest = 7+15 = 22