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Nine Marbles (Posted on 2014-09-20) Difficulty: 3 of 5
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?

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution A simpler solution | Comment 3 of 4 |
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.

  Posted by Steve Herman on 2014-09-20 18:54:15
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