Question

There are 5 consecutive odd integers. The sum of the three least is 3 more than the sum of the two greatest. Find the integers.

Answer 1

Answer:
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.