2
$\begingroup$

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:

  1. 10
  2. 15
  3. 8
  4. 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?

$\endgroup$

3 Answers 3

2
$\begingroup$

Here is an easy python brute-forcing code a programmer can whip up to solve the problem:

from itertools import combinations

weights = [i for i in range(1, 20)]

for s in combinations(weights, 9):
    not_steel = [i for i in weights if i not in s]
    for b in combinations(not_steel, 9):
        if sum(s) - 90 == sum(b):
            g = [i for i in not_steel if i not in b][0]
            print(f'Weights of the steel weights: {s}')
            print(f'Weights of the brass weights: {b}')
            print(f'Weight of the gold weight: {g}')

Output:

Weights of the steel weights: (11, 12, 13, 14, 15, 16, 17, 18, 19)
Weights of the brass weights: (1, 2, 3, 4, 5, 6, 7, 8, 9)
Weight of the gold weight: 10

$\endgroup$
2
  • $\begingroup$ Good job! You have explained in a easy way, I understand it easier, thank you! $\endgroup$
    – AntsPiano
    Oct 29, 2020 at 17:30
  • $\begingroup$ @Jingbothedude I'm assuming this is not sarcasm so thanks! $\endgroup$ Oct 29, 2020 at 18:18
2
$\begingroup$

Follow the same path you already got, but do not assume; observe and deduce instead.

Consider possible groups of nine distinct weights. The maximum possible total weight is $19+18+\dots+11 = 135$. And the minimum possible total weight is $1+2+\dots+9 = 45$.

Now observe that the maximum difference between two groups is equal to the maximum possible minus the minimum possible, which is $135 - 45 = 90$. And the description says the difference is exactly 90, so we can deduce that the group of steel weights must have the maximum possible total weight, and the group of brass must have the minimum possible.

Then, again observe that there is only one way to get the maximum possible total weight and the minimum possible total weight. So we can deduce that the steel ones must be $11 \dots 19$ and the brass ones must be $1 \dots 9$. This leaves the only one choice for the gold one: 10.

$\endgroup$
1
  • $\begingroup$ In the end it seemed that it is the quickest and logical way to obtain the weights. It looks that this is the only possible combination to get the $45\,g$ and the $135\,g$. The thing as I mentioned you stop when you reach to nine of those weights. $\endgroup$ Nov 8, 2020 at 3:27
1
$\begingroup$

1 Get the total weight : 190

2 Since we need to choose from the 4 given choices: Get the total weight for brass in those 4 cases (steel could also be used but results in higher numbers, what typically is a bit more difficult)
1 $(190-90-10)/2 = 45$
2 $(190-90-15)/2 = 42.5$
3 $(190-90-8)/2 = 46 $
4 $(190-90-12)/2 = 44 $

3 Since these numbers are pretty low: Get the minimum possible: 45 (by using 1g..9g) option 2 already failed since not integer
option 4 fails since too low
option 1 works (without further adaptations, since 10g is not used for brass)
option 3 also fails: Since the gold one is 8g, and thus cannot be brass, here the 8g cannot be used to get the minimum possible for brass, it needs to be replaced by at least the 10g; making the minimum brass total 47g

Alternative proof ascertaining uniqueness:

If the 9 brass weights are the smallest (1g..9g), and the steel weights the heaviest (11g..19g), the weight difference is 90 (11g-1g=10g; 12g-2g=10g etc., thus 9*10)
Since any other configuration means less weight difference , this is the only solution, and the gold one must be the unused one: 10g

$\endgroup$
2
  • $\begingroup$ Where and how did you obtained the weights of 10, 15, 8, 12?. I don't get very well why 8g doesn't work?. What do you mean that it has to be replaced by 10g?. How do you obtained 47g to be the minimum brass total? Why did you divide by two?. Can you explain this part on how did you obtained the weights for the brass? How did you come to the conclusion that for brass there are only 4 cases?. The way how I think you got the division by two is by making an equation such as: $s+b+g=190$, since $s=b+90$, then $2b+90+g=190$, hence $b=\frac{1}{2}(190-90-g)$ $\endgroup$ Oct 28, 2020 at 23:16
  • $\begingroup$ But that's it how far I went with understanding your analysis. From here based on the weighs means that g can be either 1,2..up to 19. How did you constrained the value of g to 10, 15, 8 and 12?. Can you include the rationale which you used to get those numbers?, I'm stuck with that part. $\endgroup$ Oct 28, 2020 at 23:20

This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .