SOLUTION: This is a problem from my math text.
Find all sets of three consecutive multiples of 3 whose sum is between -134 and -109.
This is how I set it up:
-134 < 3n + 3(n+1) + 3(n
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Find all sets of three consecutive multiples of 3 whose sum is between -134 and -109.
This is how I set it up:
-134 < 3n + 3(n+1) + 3(n
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Question 250147: This is a problem from my math text.
Find all sets of three consecutive multiples of 3 whose sum is between -134 and -109.
This is how I set it up:
-134 < 3n + 3(n+1) + 3(n+2) < -109
But it can't be right because the solution is very messy. Can you set this up for me?
Thanks! Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! I am not sure I know how to do this mathematically but your three consecutive multiples of three have to be n, n+3 and n+6
So that we have -134>3n+9<-109
33+36+39=108 too low
36+39+42=117 lowest triplet
39+42+45=126 highest triplet
42+45+48=135 too high
