The problem is as follows:

A kid has five marbles. These marbles have different weights and the child weighs them in pairs in all possible ways. He records the weights in his notebook. These are the results: 10g, 12g, 13g, 14g, 15g, 16g, 17g, 18g, 20g and 21g. Using this information, what is the weight of the lightest marble?

The choices given in my book are as follows:

  1. 4g
  2. 3g
  3. 2g
  4. 5g

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.

I'm having trouble with accounting for the weights being paired.

So far the only thing I could noticed is that if I were in that situation I would label the marbles as:


All the combinations without repetitions (which I'm assuming is the intended meaning) would be:


which indicates the 10 pairs given in the problem. But that's it. I don't know if this can be used to get an answer.

Does an easier way to make some equations or get an answer exist?

Another way would be to build a set of 10 equations with 10 unknowns. But I don't think that would be the intended method of solution. Even if such set is made, which would correspond to which weight?

Can someone help me here? Does a way to simplify this situation exist?

Please only give detailed, step-by-step solutions. No matter how I look at this question, I get tangled with equations.

  • 3
    $\begingroup$ To start: You have indeed 10 pairs, but 12 different weights, so I'd say the problem as stated has no solution $\endgroup$
    – Retudin
    Oct 28, 2020 at 19:48
  • $\begingroup$ @Retudin I'm very sorry. I did transcribed this in a rush. As it is featured now it has the right weights. As you notice there are only 10 of those weights. As it stands now, can it be solved using a simple analysis?. $\endgroup$ Oct 28, 2020 at 23:26
  • $\begingroup$ @Bubbler I have spotted one or two errors here and there, but I think the errors are mainly caused because I typed them in a rush. But if it is referred because of the degree of difficulty I think it is that challenging part which makes me keep trying more and more. $\endgroup$ Oct 28, 2020 at 23:33
  • $\begingroup$ @ChrisSteinbeckBell: "These marbles have different weights". Does this mean no two balls have the same weight? $\endgroup$ Oct 30, 2020 at 19:35
  • $\begingroup$ @ Chris Ssteinbeck.Why don't you respond to new answers? $\endgroup$ Nov 1, 2020 at 21:32

4 Answers 4


Using the corrected weights:

Let's call the weights $A, B, C, D, E$ in increasing order (so $A<B<C<D<E$). Note that no two marbles can have equal weights, since if e.g. $A=B$, you'll get two pairs having the same weight as in $A+C=B+C$.

As you already figured out, the 10 combinations are $AB, AC, AD, AE, BC, BD, BE, CD, CE, DE$. We don't quite know which weight corresponds to which combination, but we do know that each weight corresponds to exactly one combination, so we know that $$10+12+13+14+15+16+17+18+20+21$$ is the same as $$(A+B)+(A+C)+(A+D)+(A+E)+(B+C)+(B+D)+(B+E)+(C+D)+(C+E)+(D+E)$$ in some order.

Simplifying the latter sum gives $4(A+B+C+D+E)$, so we get the equation $$4(A+B+C+D+E)=156\\A+B+C+D+E=39$$

Also, we can find out that $AB$ is the smallest, $AC$ is the next smallest, and $DE$ is the largest, so $$(A+B)+(A+C)+(D+E)=10+12+21=43$$ Subtraction between the two equations gives $A=4$. Therefore the correct answer is Choice 1.

In similar ways, you can identify all five marbles' weights: $$A=4, B=6, C=8, D=9, E=12$$

  • $\begingroup$ So... what is the answer? Just askin $\endgroup$
    – AntsPiano
    Oct 29, 2020 at 0:01
  • $\begingroup$ @Bubbler You're nice!. But I'm wondering, why did you assumed an increasing order the marbles?. Would it work assuming the reverse order, I mean, assuming $A>B>C>D>E$? I was very stuck on why did you used this equation $(A+B)+(A+C)+(D+E)=10+12+21=43$ it turns out that this was chosen strategically to subract the weight of all the marbles to obtain the lesser weight. Maybe was this the reason why you did used an increasing order of marbles instead of a decreasing order in their weights or was just indistinct?. $\endgroup$ Oct 29, 2020 at 1:30
  • $\begingroup$ @Bubbler If you don't mind, can you add what if someone choses a reverse order?. Would it yield the same result?. $\endgroup$ Oct 29, 2020 at 1:32
  • $\begingroup$ There's no difference between assuming an increasing weights or decreasing one; I chose increasing because it felt more natural to me. If you want to assume decreasing, you can switch the variables $A, B, C, D, E$ into $E, D, C, B, A$ respectively, and solve for $E$ which is asked in the question (the lightest marble). The results should be the same, giving $E=4, D=6, \dots$. $\endgroup$
    – Bubbler
    Oct 29, 2020 at 1:34

We know that the minimum a marble can weight is 2g because of the given options. The maximum amount a marble can weigh will then be 21 - 2 = 19g.

Here is my brute-force method for those who are interested in brute-force solutions (only takes two seconds to return the output)!

from itertools import combinations as comb

weights = [10, 12, 13, 14, 15, 16, 17, 18, 20, 21]

for a in range(2, 20):
    for b in range(2, 20):
        for c in range(2, 20):
            for d in range(2, 20):
                for e in range(2, 20):
                    test = [i + j for i, j in comb([a, b, c, d, e], 2)]
                    if sorted(test) == weights:
                        print(f'A = {a}, B = {b}, C = {c}, D = {d}, E = {e}')
                        print(f'The lightest marble weighs {min([a, b, c, d, e])}g.')
                        input('Press enter to quit >>> ')


A = 4, B = 6, C = 8, D = 9, E = 12
The lightest marble weighs 4g.
Press enter to quit >>> 

  • $\begingroup$ I'm sorry. I modified the question because it was incorrectly typed. Now it is right. As it is now, can it be solved?. You may want to update your answer to reflect the changes I made in the question. $\endgroup$ Oct 28, 2020 at 23:35
  • $\begingroup$ @Bubbler, apparently this question can have a solution, its just that I got a typo and once I noticed I had to make the edit so it will update the question. $\endgroup$ Oct 28, 2020 at 23:36
  • $\begingroup$ @ChrisSteinbeckBell Updated! $\endgroup$ Oct 29, 2020 at 0:27

I say the minimum weight is 4.

Since we have five different weights and we take two at a time, then if we apply the formula of combinatorics with no repetitions, the number of combinations is ten. Since the question assigns the weights the letters A,B,C,D,E we obtain the following ten combinations.

AB  BC   CD   DE

AC  BD   CE 



Let AB be the minimum, AB=10, and DE=21 the maximum. If we set A=4 then, according to the given facts, we have

AB 10-4=6 so B=6

AC 12-4=8 so C=8

AD 13-4=9 so D=9

DE 21-9=12 SO E=12

So from the weights 4, 6, 8, 9, 12 we can obtain all ten combinations.


The lightest two marbles weigh 10g, and the next lightest combo 12g, so the weight difference between second and third lightest is 2g.
Similarly the weight difference between second and third heaviest is 1g.
so we have a combo x(<y),y, y+2g ,y+3g, z(>y+3g)
With this, we know that the pairs with the lightest are 10g, 12g and 13g; and with the heaviest 18g, 20g and 21g.
The pairs with only the middle 3 weigh 2y+2,2y+3,2y+5
Fitting that on 14,15,16,17 gives y=6 i.e. weights of 6g,7g and 9g (using up 14g,15g,17g)
The lightest must then be 4g and the heaviest 12g which together indeed yields the missing weight 16g


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