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Weighing Jelly Babies Puzzle - Solution
The Puzzle:
Real jelly babies have a mass of 10 grams, while imitation jelly babies have a mass of 9 grams.
Spike has 7 cartons of jelly babies, 4 of which contain real jelly babies, the others imitation.
Using a scale only once (and minimum disturbance of the boxes contents!), how can Spike determine which cartons contain real jelly babies?
Our Solution:
Remove a DIFFERENT number of jelly babies from each box, weigh the boxes, then figure out which combination would produce that weight loss.
The best method is to remove 0, 1, 2, 4, 7, 13, and 24 jelly babies from the boxes. 51 total.
If they were all genuine, you would have removed 51 x 10g = 510g, but not all boxes are genuine so the weight will be less.
The trick is in choosing the right numbers to remove, so that all possible combinations will be unique.
Start at 0 and then choose each new number to be greater than any previous, because otherwise the sums will not be unique, but also the sums it made when paired with any previous number have to be distinct from all previous pairs. The number 0, 1, 2, 4, 7, 13, and 24 work well.
So, upon weighing, the weight will be less than 510g.
If it is only 3g lighter the combination must be 0, 1 and 2
If it is 5g lighter the combination must be 0, 1 and 4
And so on.
If you are curious, the possible combinations are:
3g = 0 + 1 + 2
5g = 0 + 1 + 4
6g = 0 + 2 + 4
7g = 1 + 2 + 4
8g = 0 + 1 + 7
9g = 0 + 2 + 7
10g = 1 + 2 + 7
11g = 0 + 4 + 7
12g = 1 + 4 + 7
13g = 2 + 4 + 7
14g = 0 + 1 + 13
15g = 0 + 2 + 13
16g = 1 + 2 + 13
17g = 0 + 4 + 13
18g = 1 + 4 + 13
19g = 2 + 4 + 13
20g = 0 + 7 + 13
21g = 1 + 7 + 13
22g = 2 + 7 + 13
24g = 4 + 7 + 13
25g = 0 + 1 + 24
26g = 0 + 2 + 24
27g = 1 + 2 + 24
28g = 0 + 4 + 24
29g = 1 + 4 + 24
30g = 2 + 4 + 24
31g = 0 + 7 + 24
32g = 1 + 7 + 24
33g = 2 + 7 + 24
35g = 4 + 7 + 24
37g = 0 + 13 + 24
38g = 1 + 13 + 24
39g = 2 + 13 + 24
41g = 4 + 13 + 24
44g = 7 + 13 + 24
The best method is to remove 0, 1, 2, 4, 7, 13, and 24 jelly babies from the boxes. 51 total.
If they were all genuine, you would have removed 51 x 10g = 510g, but not all boxes are genuine so the weight will be less.
The trick is in choosing the right numbers to remove, so that all possible combinations will be unique.
Start at 0 and then choose each new number to be greater than any previous, because otherwise the sums will not be unique, but also the sums it made when paired with any previous number have to be distinct from all previous pairs. The number 0, 1, 2, 4, 7, 13, and 24 work well.
So, upon weighing, the weight will be less than 510g.
If it is only 3g lighter the combination must be 0, 1 and 2
If it is 5g lighter the combination must be 0, 1 and 4
And so on.
If you are curious, the possible combinations are:
3g = 0 + 1 + 2
5g = 0 + 1 + 4
6g = 0 + 2 + 4
7g = 1 + 2 + 4
8g = 0 + 1 + 7
9g = 0 + 2 + 7
10g = 1 + 2 + 7
11g = 0 + 4 + 7
12g = 1 + 4 + 7
13g = 2 + 4 + 7
14g = 0 + 1 + 13
15g = 0 + 2 + 13
16g = 1 + 2 + 13
17g = 0 + 4 + 13
18g = 1 + 4 + 13
19g = 2 + 4 + 13
20g = 0 + 7 + 13
21g = 1 + 7 + 13
22g = 2 + 7 + 13
24g = 4 + 7 + 13
25g = 0 + 1 + 24
26g = 0 + 2 + 24
27g = 1 + 2 + 24
28g = 0 + 4 + 24
29g = 1 + 4 + 24
30g = 2 + 4 + 24
31g = 0 + 7 + 24
32g = 1 + 7 + 24
33g = 2 + 7 + 24
35g = 4 + 7 + 24
37g = 0 + 13 + 24
38g = 1 + 13 + 24
39g = 2 + 13 + 24
41g = 4 + 13 + 24
44g = 7 + 13 + 24