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?

Label the marbles 1 to 9 from lightest to heaviest.
Start by figuring the min and max for each friend.
Amin and Cmin are both equal to the sum of the smallest 3 marbles:  39
Bmin is twice Amin.
(A+B)max is therefore the total 398 - 39 = 359
Since A=2*B Bmax = Floor(359/3) = 119; Amax = 238.
Cmax is the Total minus (A+B)min

So far we have:
     min   max
A    78    238
B    39    119
C    39    281

If Bmax is 119, then B can not have marble 8 or 9
So Bmax would be marbles 5, 6 and 7:  87

     min   max  marbles
A    78    174
B    39     87  !8 !9
C   137    281

Looking at Amax, I see that marbles #1 + #2 + #9 = 177,
so A can not have marble 9, nor can B therefore C does.
So Cmin is 177, and
(A+B)max is therefore the total 398 - 177 = 221

     min   max  marbles
A    78    146  !9
B    39     73  !8 !9
C   177    281  #9

The total is 2mod3, (A+B) is 0mod3, and so C is 2mod3.
Marble 9, which C has, is 1mod3 so the other two marbles held by C have a total which is 1mod3.
Now we have a bit of luck because each marble, mod 3 is as follows:
10    1
13    1
16    1
17    2
19    1
22    1
46    1
101   2
154   1

Since C owns the heaviest marble, the only way he can find two others whose sum is 1mod3 is for those 2 marbles to be #4 and #8 (the only ones that are 2mod3).
So the total held by C is 17+101+154=272 which is 2mod3
Therefore A total is (Total - C)/3 = 42;  and B total = 84.

                         Total
Ade     16   22    46     84
Bill      10   13    19     42
Cade   17  101  154   272

Edited on September 20, 2014, 9:24 am

Posted by Larry on 2014-09-20 09:22:14