In ten barrel face number riddle, rearrange ten barrels in a pyramid so that sum of face numbers of four barrels on each outer side is equal and minimum.
The Riddle
In a storehouse, ten barrels are kept in the form of pyramid—four at the bottom, three on top of the four, two on top of the three and one at the very top. The barrels are numbered 1 to 9. The tenth barrel is not numbered. The barrels are so arranged that sum of barrel numbers at the base and the two sides of the pyramid all are 16.
The 10 barrels in the form of a pyramid:
Rearrange the barrels in such a way that the sum of the barrel numbers at the base and on the two sides of the pyramid are equal and the least.
Recommended time to solve: 30 minutes.
Puzzle courtesy: Henry Ernest Dudeney, the legendary puzzlist from UK.
This looks small, but is not easy. We assure you though, if you try with reason and serious intent, you should be able to solve it with no trouble.
Solution to ten barrel face number riddle
Briefly analyzing the riddle take the first strategic decision,
Strategy: You will analyze the combinations of the digits on the three sides of the pyramid together. The goal will be to analyze how the sum of each 9 digit combination on the three sides behave. The tenth digit at the center won’t make any contribution to the total and will be ignored.
This makes sense, as analyzing the sum of digits on the three sides separately will be too uncertain and complicated.
In the following schematic, the pyramid with the barrel faces are numbered for ease of understanding.
In the puzzle configuration, the number 7 is at the center and does not contribute to the three sums of the numbers on three sides of the outer triangle (connecting centers of the circles).
The unnumbered face of the barrel is now numbered as 0. Zero will be its contribution to the sum of the face numbers on the triangle side containing the barrel.
According to the strategy adopted, we will first add up the ten digits from 0 to 9. The sum is 45.
The first key pattern of this sum of ten digits identified,
Key pattern 1: The digit placed at the center will be out of consideration for summing up the numbers on the sides of the outer triangle.
So the total 45 of 9 digits on the three sides of the triangle is reduced by the number at the center (as it is the 10 digit total reduced by the digit at the center). In puzzle configuration, the sum of the 9 digits on the periphery is thus, 45 - 7 = 38 as digit 7 is at the center.
First decision to take, which of the nine digits you should place at the center thus reducing the total sum by that number!
Reasoning: As the sum of three face numbers on each side of the triangle must be the least, their sum (sum of peripheral 9 face numbers) must also be the least. This happens only if the largest face number 9 is left at the center, so that the total 45 of ten face numbers is reduced to the greatest extent.
So first key decision taken,
Key Decision 1: Barrel with 9 on its face will be at the center.
Sum of rest 9 digits is reduced now to 36.
Next important pattern discovered is,
Key Pattern 2: With nine digits on the three sides of the triangle, the sum of each side is formed first and then the three sums of the three sides added together. This total is NOT of 9 DIGITS, but of 9 + 3 = 12 digits. Each corner digit is added twice for the two sides containing it. One sum of three corner digits is extra.
For example, in the puzzle, the digit 1 is added once for the sum of side 1-8-4-3 and a second time for the sum of 1-0-9-6. Likewise, for the other two corner digits 3 and 6.
So the sum of nine peripheral digits 38 (45 less 9) is jacked up by the sum of three corner digits 1 + 3 + 6 = 10 to 48 as the sum of the three sums of the sides of the triangle. Total of each side is 16 and 3 times 16 is 48. The idea of double contribution of corner digits is corroborated.
How do you use this important discovery?
Answer is easy. For the sum of each triangle side, and sum of the three sides to be the least, make sure that the smallest digits are at the corners. This will reduce the total of the three sides to the largest extent.
Key Decision 2: You must place the smallest three digits 0, 1 and 2 at the three corners.
The schematic of the corner digit plan:
As 9 is at the center, the sum of the rest nine digits is reduced to 36. So the critical conclusion on the smallest sum of each triangle side is,
Critical Conclusion 1: Each of the three periphery sums must be larger than one-third of 36 equal to 12, because this total of sums of three sides will be increased by the sum of corner digits 1 + 2 = 3 to 39.
It follows,
The smallest sum of each triangle side will be 13, one-third of 39.
One step more to the solution. The actual barrel arrangement for the solution to be decided.
Barrel arrangement for least sum of four face numbers on each side of the triangle
How to get the sum of each triangle side as 13? It is easy.
For the side with digits 0 and 1, total of two barrel face numbers placed at the rest two positions must be 12. In the same way, for the side with (0, 2) sum of rest two digits in the side must be 11 and for the side with (1, 2), it must be 10.
So,
Critical conclusion 2: Split rest six digits 3, 4, 5, 6, 7, 8 into three pairs so that sum of the three pairs are 10, 11 and 12 for triangle sides with corner digits (1, 2), (0, 2) and (0, 1) respectively to make sum of four barrels on each peripheral side 13.
You can form 10 as the sum of two digits out of six as:
3 + 7 = 10 and
4 + 6 = 10.
These are the only two ways.
Exploring solution for the sum 3 + 7 = 10: First solution to ten barrel face number riddle
With sum 10 of digits 3 and 7, explore how the sums 11 and 12 can be formed by two pairs of digits out of the four digits 4, 5, 6 and 8.
Easy to find: 5 + 6 = 11 and 4 + 8 = 12, the only possible way.
We have got our first solution.
The three four-digit sets in this solution, anti-clockwise, are:
0, 4, 8, 1: sum 13
1, 3, 7, 2: sum 13
2, 5, 6, 0: sum 13.
Verify the first solution from the schematic.
Second solution to ten barrel face number riddle
Analyze the second possibility of sum 10 as: 4 + 6 = 10.
As before, split the rest four digits 3, 5, 7 and 8 into two pairs with sums 11 and 12. Can we do this?
Yes, in only one way,
3 + 8 = 11 and 5 + 7 = 12.
This creates the second solution, and that is all. The second solution is,
0, 5, 7, 1: sum 13
1, 4, 6, 2: sum 13
2, 3, 8, 0: sum 13.
Verify the barrel arrangement of the second solution from the schematic.
The systematic method ensures that these are the only two solutions.
The puzzle looks simple but poses a hard challenge. It is solved mainly by strategy and reasoning, not math.
The process of solution formed a systematic method that can be used for exploring this type of puzzles further.
Bonus puzzles on Ten barrel face number riddle
Can you find out any arrangement of the ten barrels for peripheral sum 14 or 15?
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