SOLUTION: the sum of three consecutive odd intergers is 17 less than four times the smallest integer

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Question 1125067: the sum of three consecutive odd intergers is 17 less than four times the smallest integer
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Consecutive odd integers
n, n+2, n+4


The description
n%2B%28n%2B2%29%2B%28n%2B4%29=4n-17
-
3n%2B6=4n-17
6%2B17=n
n=23


The consecutive odd integers
23, 25, 27

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The algebra is often easier in a problem like this if you let x be the middle of the three integers. If you are good with mental calculations, you might be able to solve the problem without writing out all the steps.

Here is how it might go (see if you can do all these steps mentally....)

Because x is the middle of the three consecutive odd integers, the sum of the three integers is now 3x.

That sum is 17 less than 4 times the smallest integer, which is x-2:

3x+=+4%28x-2%29-17
3x+=+4x-8-17+=+4x-25
x+=+25

The middle number is 25; the three numbers are 23, 25, and 27.