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 20140920 09:22:14 