The problem is as follows:
Vincent has 19 different calibrated weights whose weights are 1g, 2g, 3g, and so on up to 19g. Nine of them are made from steel, nine are made from brass, and the remaining one is made from gold. Vincent knows that the total weight of all the steel weights is 90g more than the total weight of all brass weights. Using only these clues, find the weight of the gold weight.
The choices given in my book are as follows:
- 10
- 15
- 8
- 12
For reference I found this problem in my collection of puzzles book Reason and Logic. From the style I believe it is an adaptation from the contents found in Martin Gardner's 50's book on Recreational Puzzles.
The thing with this problem is I don't know how to solve it in logical, non-guessing way.
What I did was assume that in order to get the heaviest weight I could add the heavier weights as follows:
$19+18+17+...$
But that's the part where I got stuck. Where to stop? My conclusion was, since it mentions that 9 of those steel weights are 90g more than the brass ones then it would meant that:
$19+18+17+16+15+14+13+12+11$
It can be expressed as:
$t_n=11+(n-1)1=10+n$
Then the sum would mean:
$$\sum ^{9}_{i=1}(10+n)= \sum ^{9}_{i=1} 10 + \sum ^{9}_{i=1}n=90+\frac{9\times 10}{2}=90+45=135$$
Then this would meant that the other group must be 45g.
But what sort of combination will yield this?.
Then I assumed that it meant the other end will yield the lesser weight possible (referring to the brass weights) and with those being 9. Hence:
$1+2+3+4+5+6+7+8+9=45g$
Therefore the only remaining weight will be 10g.
Hence the gold weight will be assigned that weight. So it must be 10g. I'm assuming that's the answer. Upon checking the answers sheet it checks.
But again, I'm not very happy with my solution path. Does an easier/more intuitive way to solve this riddle exist?
I was confused on how to assume which combination will be assigned to each group of weights. There isn't any reason to specifically say that the group between 19 to 11 will generate 90g plus something. Would some other combinations work? i.e maybe between 18 to 10. That's the part which I found confusing.
In the end, the only logic which I could find was if I were to use that combination maybe it would yield a contradiction. But is there any way to prove that other combinations will cause contradictions?
Can someone guide me with a solution analyzing all the possible cases? As I indicated, it took me some time to realize which combination would get the answer. Does a straightforward way to approach this puzzle exist?