Question

There are five consecutive odd integers. The sum of the three smallest is 3 more than the sum of the two largest. Find the integers. 15 points

Answer 1

11, 13, 15, 17, 19

Explanation:

I am doing this in my RSM homework, just saying this is correct. I guessed and checked to do this.

Answer 2

11 13 15 17 19

Explanation:

We will solve this word problem using 2k+1 which is one of the general forms of an odd integer.

Let 2k+1 be the first odd integer.

(2k+1) + (2k+3) + (2k +5) = (2k+7) + (2k +9) + 3 (the +3 is to accommodate for the left side being 3 more than the right)

Solve for k

6k + 9 = 4k + 19

6k - 4k + 9 = 4k-4k + 19

2k + 9 = 19

2k + 9 - 9 = 19 - 9

2k = 10

2k/2 = 10/2

k = 5

To get the first digit in the series plug in 5 for k

2k+1

2(5) + 1

10 + 1 = 11

So our series is 11, 13, 15, 17, 19

Let's test it.

11 + 13 +15 = 39

17 + 19 = 36

So we are correct, the first 3 integers are 3 more that the sum of the last 2.