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The following pentagon is composed of 10 discs. Each disc must be filled with one number. The first 10 even positive numbers are allowed. The sum of the numbers to be placed in each side of the pentagon must be the same and the maximum possible. Find the sum of the numbers adjacent to the discs filled with the numbers 2 and 18.

The figure is shown below:

Sketch of the problem

The choices given are:

  1. 54
  2. 60
  3. 50
  4. 64

I tried several ways to fill up this pentagon but did not find a clear solution. Does a strategy to solve this kind of puzzle exist? I'm stuck at the very beginning. I believe this requires mathematics. Please try to include some visual aid and the most detailed way to solve this problem.

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    $\begingroup$ Need some clarification: In the question, you said to fill the pentagon with first ten positive even numbers, but at the end of first paragraph you ask the pentagon to be filled by only 2 and 18? $\endgroup$
    – SamRoy
    Oct 13, 2019 at 13:42
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    $\begingroup$ @SamRoy Fill the circles with each of the even numbers $2$ to $20$ as described. Then take the sum of the numbers that are adjacent to the numbers $2$ and $18$ in the filled diagram. $\endgroup$ Oct 13, 2019 at 18:31

2 Answers 2

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The puzzle asks for a maximal sum for the numbers on the edges, so...

The sum of all five edges will include each number at a vertex twice, once for each edge that it is part of. We need to put the largest numbers at the vertices in order to maximize the sum. This will give us a maximum total sum of $2(20+18+16+14+12)+10+8+6+4+2=190$. Dividing the among the five edges gives us a maximum equal edge sum of $\frac{190}{5}=38$. In order to obtain equal sums on all edges, we need to balance the high and low numbers. Starting with $20$ at a vertex, we add $2$ and $4$ to either side of the $20$. Given the target sum of $38$, we complete the first two edges as $20+2+\color{blue}{16}=38$ and $20+4+\color{blue}{14}=38$. Now consider the other two vertices. We need to put $12$ and $18$ in those discs. If we put $18$ on the edge with $16$, we would need $16+4+18=38$, but we have already placed the $4$. So we place $18$ with $14$ and $12$ with $16$ which gives us $16+10+12=38$ and $14+6+18=38$, with the final number, $8$, we have $18+8+12=38$ and the diagram is complete. The numbers adjacent to $2$ and $18$, with their sum, are $16+20+8+6=\color{blue}{50}$

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  • $\begingroup$ Nice approach. +1 $\endgroup$
    – SamRoy
    Oct 15, 2019 at 4:00
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    $\begingroup$ Placing 2 and 4 next to 20 is a bit arbitrary. Why not 2 and 6? Couldn't it also work? You could have started more logically by noting that 2 can only be in a line with 20 and 16 and 10 only with 12 and 16. $\endgroup$
    – Florian F
    Apr 10 at 10:13
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    $\begingroup$ @FlorianF Placing 2 and 6 next to 20 puts 16 and 12 on their respective vertices, leaving 14 and 18 on the remaining side needing the 6 again for 38. Your logic leading to 20-2-16-10-12 leaves only one place for 18, proving the solution unique. $\endgroup$ Apr 10 at 12:19
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The answer is:

50

Because the filled-up pentagon looks like:

enter image description here

The sum in each edge is 38. This is the maximum I got after several trial & errors.

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  • $\begingroup$ Indeed, this is correct. See my answer for details on how to arrive at this result more directly. $\endgroup$ Oct 13, 2019 at 19:06

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