

The product N of three positive integers is 6 times their sum , and one of the integers is the sum of the other two.
Find the sum of all possible values of N . product N of 3 positive integers A x B x C = N is 6 times their sum , = 6 ( A + B + C ) and one of the integers is the sum of the other two. A + B = C Find the sum of possible values of N A x B x C = N = 6 ( A+ B+ C ) A + B = C so A+ B = A+ B A x B = AB AB ( A+ B) = N = 6 ( A + B + A + B ) ...............A+ B is substituted for all C's see A+B = C AB ( A + B ) = 6 ( 2A + 2B ) = N AB ( A + B ) = 12 ( A + B ) = N divide both sides by ( A + B ). AB ( A + B ) / ( A + B ) = 12 ( A + B ) / ( A+ B ) = AB = 12 = N ..........................Product of 3 + integers.........6 ( A+ B + C) A + B = C ....................AB(C) = N.................. 1 + 12 = 13 ........ 12 X 13 = 156........... 6 ( 1 +12 +13 ) = 6 x 26 = 156 2 + 6 = 8 ..........2 x 6 X 8 = 96............6 ( 2 + 6 + 8 ) = 6 x 16 = 96 3 + 4 = 7 ..........3 x 4 x 7 = 84 ............6 ( 3 + 4 + 7 ) = 6 x 14 = 84 Find the sum of possible values of N 156 + 96 + 84 = 336 




Pemdas





N=6
A=1 B=2 C=3 Not great at math, but my best guess Yawni 




Oh these are great
I really liked your birdhouse with the picket fence on it too (I got 3 sets of possible numbers for A,B,C  is that what you got? ) 




very good Sam guess I will repost it
video of the math problem http://www.khanacademy.org/video/200...etition%20Math 




Maybe people are just shy about like, oh geez, did I do it okay? Is my answer right? Is it part right? Sort of?
They're getting harder pretty fast! But nice to be able to try and stuff 




I can do a range
maybe I should back pedal If nothing else I want to raise players math skills to the point they can do all the math behind poker 


