Three friends Ade, Ben and Cade have 9 marbles in front of them. Their weights are as follows:
154, 16, 19, 101, 10, 17, 13, 46 and 22 grams.
Each of the three friends takes 3 marbles. Ade's bunch weighs precisely twice as much as Bill's bunch.
What is the total weight in Cade's bunch?
Full credit to Larry for solving this (nice work)!
Using his insights, a simpler solution is available.
a) The total weight is 398, which equals 2 mod 3
b) A + B equals 0 mod 3, so C equals 2 mod 3
c) All of the weights equal 1 mod 3, except for 101 and 17, which equal 2 mod 3
d) The only way that C can equal 2 mod 3 is if C has 101, 17 and one other weight.
e) B clearly cannot have 154, because A's weight would then be impossibly high. So B's maximum weight is 46 + 22 + 19 = 87.
f) A's maximum weight is therefore 87*2 = 174, so A cannot have 154 either, as his bunch would then weight over 174.
g) Therefore Cade has 154, 101 and 17, for a total of 272.
A simpler solution
