Five friends are sharing information about the marbles they have attempting to derive the total amount of marbles they own without disclosing to others the amount each own. They decide to do it by having two disclosing the sum of their marbles to the interested individual that is not part of the group:

John and Chuck have a total of 10

Nick and Chuck have a total of 13

Nick and Paul have a total of 15

Paul and John have a total of 12

How many they have total? Can you derive how many each one of them owns? When they meet Roger they add him to the group by modifying the last sum as follows:

Paul and Roger have a total of 14

Roger and John have a total of 8

What is the total of marbles the 5 friends have? Are they still preventing others from knowing how many marbles each of them have?

Could you explain the difference?

  • $\begingroup$ If you did want to find the total without revealing the components, the algorithm is that the first person adds a random number to their number and tells the next person secretly. They add their number and pass it secrets on to the next. This goes on until they have all added their number and it is secretly told to the first person. The first person subtracts the random number and announces the total! I guess that's how the pairs did it! $\endgroup$
    – Dr Xorile
    Dec 13, 2018 at 23:07
  • $\begingroup$ I did not want to do it - I believe I know how to encrypt. It was just interesting to see that sometimes very similar problems are not leading to having the same solution. $\endgroup$
    – Moti
    Dec 15, 2018 at 0:46

2 Answers 2


For the first part

In the first example, denote the amount of marbles each friend has by $J, C, N$ and $P$ (corresponding to the first letter of their name). Then, we have $$ 2(J+C+N+P) = (J+C)+(N+C)+(N+P)+(P+J) = 10+13+15+12=50 $$ and so the total number of marbles is $$J+C+N+P = 25 $$ It looks like we should be able to derive all their values but notice that the fourth equation is redundant. For example, using the first three equations, we have $$ J+P = (J+C) - (C+N) + (N+P) = 10 - 13 + 15 = 12$$which is exactly the fourth equation. Hence, we really have three equations in four unknowns and there is not a unique solution. This means that the number of marbles an individual has is kept secret.

For the second part,

This time let the number of marbles Roger has be $R$. Then, we have $$ 2(J+C+N+P+R) = (J+C)+(N+C)+(N+P)+(P+R)+(R+J) = 10+13+15+14+8 = 60$$ and thus, the total is $$ J+C+N+P+R = 30 $$ This time we really do have five equations in five unknowns. For example, we can work out that $$J-R =(J+C)-(C+N) +(N+P)-(P+R) = 10-13+15-14 = -2$$ and together with $ J+R = 8$ we find that $J=3,\,\, R=5$ This gives us the rest $$C=7,\,\,N=6,\,\,P=9 $$

The difference

In the first case, there is an inherent symmetry in the system which makes one of the equations redundant. In the second case, there is no such symmetry. The number of marbles can always be determined when the number of participants is odd but not when even.


After Roger's arrival, we know Paul has 6 more marbles than John. So, Paul=9, John=3, Nick=6 and Chuck=7. Total 25. Roger has 5 marbles. Total 30.


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