Question #cb56b

2 Answers
May 19, 2017

{2n+1, 2n+3, 2n+5}={11, 13, 15}{2n+1,2n+3,2n+5}={11,13,15}

Explanation:

If you have three, consecutive odd integers, you could write it like this
{2n+1, 2n+3, 2n+5}{2n+1,2n+3,2n+5} where nn is an integer

The 2n2n term is guaranteed to be even. So adding an odd value (i.e., 1, 3, and 5) ensures it will be odd.

The smallest odd is 2n+12n+1. The largest is 2n+52n+5. The problem tells us that

2(2n+1)=2n+5+72(2n+1)=2n+5+7
4n+2=2n+124n+2=2n+12
2n+2=122n+2=12 (subtract 2n2n from both sides)
2n=102n=10 (subtract 22 from both sides)
n=5n=5

Plugging n=5n=5 back into our three consecutive integers gives
{2n+1, 2n+3, 2n+5}={11, 13, 15}{2n+1,2n+3,2n+5}={11,13,15}

May 19, 2017

Let xx be the smallest of the three consecutive odd integers. Then the three consecutive odd integers are:
x, x+2, x+4x,x+2,x+4

2x = 7+(x+4)2x=7+(x+4)
Left side of the equation comes from "Twice the smallest"
Right side of the equation comes from "seven more than the largest"

2x=x+112x=x+11

x=11

Since we have xx now, the three consecutive odd integers are:
11, 11+2, 11+411,11+2,11+4

11, 13, 1511,13,15