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The above picture represents the drums stocked with wine at the shop. Each number represents the quantity in the same unit.

At the end of the day, the last remaining drum is retained by the store owner since it contains red wine, while the rest contained white. While going through the sales record of the day, the owner realized the following, "The first buyer received three times the quantity as much as the second while the third received twice as much as the second."

Can you identify which drum contained the red wine?

As a bonus you can also mention the quantity purchased by each buyer.

Addendum : To make the solution unique "The number of drums sold to each were the same".

Note : I am more interested in the methodology used to arrive at the solution. Brute force may help, but the problem should be solvable using only pencil and paper. Derived from a century old puzzle.

  • $\begingroup$ Is your answer unique? I.e. can you deduce a single drum left unsold (with no possibility of any other) from the given clues? So far, I can get it down to (one of) the 3, 9, or 15 unit drums. $\endgroup$
    – Lawrence
    Commented Oct 25, 2015 at 7:50
  • $\begingroup$ @Lawrence The Answer is not unique. $\endgroup$
    – kanchirk
    Commented Oct 25, 2015 at 8:47
  • $\begingroup$ @Lawrence Based on your comment, I have added a condition to make the solution unique. But both the answers are indeed correct. Sincere Apologies for this late addition. $\endgroup$
    – kanchirk
    Commented Oct 25, 2015 at 9:11

2 Answers 2


The total number of available units is $207$. The total amount of sold units must be divisible by $6$. So the number in the remaining drum must be $3+6x$. If we assume the owner didn't want to move the drums around, then the remaining drum must be one of the bottom ones. The only one matching is $15$.

To find the drums sold to each customer I started with the last customer and again drums from the bottom:

  • third customer: $10+11+13+13+17=64$
  • second customer: $9+10+13=32$
  • first customer: the rest except $15=96$


The additional requirement is easy to find, as each customer bought $7$ drums:

  • first customer: $11+12+13+13+13+17+17=96$
  • second customer: $3+3+3+5+5+6+7=32$
  • third customer: the rest except $15=64$

And it still isn't unique:

  • first customer: $11+13+13+13+15+17+17=99$
  • second customer: $3+3+3+5+5+6+8=33$
  • third customer: the rest except $9=66$
  • $\begingroup$ That is a great challenge to the original problem "Not moving drums around". You have got one of the correct answers. There is another underlying challenge I can add, to make the solution unique. "Each bought the same number of drums". $\endgroup$
    – kanchirk
    Commented Oct 25, 2015 at 9:08

Say the second buyer bought $x$ units. Then $6x$ units were sold (assuming no other customers that day). The drums sum to $207$ units. The closest multiple of $6$ to this is $204$, leaving one of the 3-unit drums.

However, assuming drums are sold in their entirety or not at all, it is possible to leave unsold any drum with $3+6x$ units for any non-negative integer $x$. This means the drum remaining could be any of the $3, 9,$ or $15$ unit drums.

New requirement: the number of drums sold were the same for each of the 3 customers (i.e. 7 each, leaving 1 for the owner).

If there is just the 3-unit drum remaining, $x=(207-3)/6 = 34$. However, after reserving one of the 3-unit drums for the owner, the smallest sum of 7 drums is 36. I.e. $x>34$. Reject.

In the other 2 cases, $x$ is 32 or 33. The smallest sum is 32, so there are possible solutions. With the given numbers, starting with the smallest sum gets you pretty close to $x$. Using that set as a starting point and swapping a number for one slightly larger, then doing the same for $2x$ gets you to a complete solution (the $3x$ combination is then determined).

Sleafar has already listed some possible solutions, so I'll just add that other solutions still remain. For example, leaving the 9-unit drum behind, the customers could have bought the following combinations: $(3,3,3,5,5,7,7), (6,8,8,10,10,11,13), (12,12,13,13,15,17,17)$.

  • $\begingroup$ Bravo !!! you got it correct. I am adding a slight modification to the problem above to make the solution unique. $\endgroup$
    – kanchirk
    Commented Oct 25, 2015 at 9:08

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