find three consecutive odd integers such that six times the sum of the first and second is 5 more than 11 times the third

Question

find three consecutive odd integers such that six times the sum of the first and second is 5 more than 11 times the third

Answer 1

Hi Allegra,

Problem: Find three consecutive odd integers such that six times the sum of the first and second is 5 more than 11 times the third

Given:

3 consecutive ODD integers

six times the sum of the first and second is 5 more than 11 times the third

Find: The 3 consecutive odd integers

Lets start with

3 consecutive ODD integers

Let n = 1st integer

The 2nd odd integer would be n + 2 (Think 1,2,3,4,5 Odd integers are 2 apart)

The 3rd odd integer would be n + 4

Now it says

six times the sum of the first and second

(6) (1st + 2nd)

6(n + n + 2)

6(2n + 2)

is means =

5 more than 11 times the third

11(3rd) + 5

11(n+4) +5

So six times the sum of the first and second is 5 more than 11 times the third

Translates to

6(2n+2) = 11(n+4) + 5

Now Solve for n

12n + 12 = 11n + 44 +5

n = 37

So the 3 consecutive odd integers are

37, 39, 41

We can check to confirm

6(37 + 39) =? 11(41) +5

456 ?= 456 Yes.. The answer checks out

Let me know if you have any more questions!