SOLUTION: find three consecutive odd integers such that twice the largest increased by 4 times the smallest is not less than 20 less than nine times the middle.

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Question 1113593: find three consecutive odd integers such that twice the largest increased by 4 times the smallest is not less than 20 less than nine times the middle.
Found 2 solutions by KMST, MathTherapy:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let the integers be
n-2 , n , and n%2B2 .
We can eliminate any even answers later.
We calculate that
"twice the largest" =2%28n%2B2%29 ,
"4 times the smallest" =4%28n-2%29 ,
"twice the largest increased by 4 times the smallest" = 2%28n%2B2%29%2B4%28n-2%29 ,
and "20 less than nine times the middle" =9n-20 .

The problem, says that 2%28n%2B2%29%2B4%28n-2%29"is not less than"9n-20 .
Because it has to be either one way or the other,
the phrase "not less than" means "more than or equal to",
so the inequality to solve is
2%28n%2B2%29%2B4%28n-2%29%3E=9n-20 .

Solving:
2%28n%2B2%29%2B4%28n-2%29%3E=9n-20
2n%2B4%2B4n-8%3E=9n-20
6n-4%3E=9n-20
6n-4-6n%2B20%3E=9n-20-6n%2B20
20-4%3E=9n-6n
16%3E=3n
16%2F3%3E=n
n%3C=16%2F3=5%261%2F3 , along with the fact that
n and n-2 are supposed to be (positive) odd integers,
means that n=5 or n=3 .
So, the "three consecutive odd integers" in the problem are
highlight%28matrix%281%2C5%2C3%2C%22%2C%22%2C5%2C%22%2C%22%2C7%29%29 or highlight%28matrix%281%2C5%2C1%2C%22%2C%22%2C3%2C%22%2C%22%2C5%29%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

find three consecutive odd integers such that twice the largest increased by 4 times the smallest is not less than 20 less than nine times the middle.
It's any 3 CONSECUTIVE ODD INTEGERS, as long as the SMALLEST <= 3.