Semi-Magical Hexagon 2

We can answer the question without obtaining the full solution... Let $a$ be the sum of the numbers in the red corner hexagons. Let $b$ be the sum of the numbers in the blue edge hexagons. Let $c$ be the sum of the numbers in the interior yellow hexagons, and let $d$ be the number in the central hexagon. Let $X$ be the magic number. Adding up the numbers in all six rows with three hexagons each, we deduce that $2a + b = 6X$ . Adding up the numbers in all six rows with four hexagons each, we deduce that $2b + 2c = 6X$ , or $b+c=3X$ . Adding up the numbers in all three rows with five hexagons each, we deduce that $a + c + 3d = 3X$ . Solving these equations we find that $a = 2X - d \hspace{1cm} b \; =\; 2X + 2d \hspace{1cm} c \; =\; X - 2d$

In this case we have $5X = a + b + c + d$ is the sum of all the numbers, which is $400$ , and hence $X = 80$ . In addition, $c \ge 5 + 8 + 9 + 11 + 13 + 14 = 60$ , and hence $60 \le c = X - 2d = 80 - 2d$ , so that $d \le 10$ .

If $d=8$ then $5+9+11+13+14+15 = 67 \le c = 80 - 2\times8 = 64$ , which is impossible. Thus $d \neq 8$ .

If $d=9$ then $5+8+11+13+14+16=66 \le c = 80 - 2\times9=62$ , which is impossible. Thus $d \neq 9$ .

Thus $d = \boxed{5}$ . In this case the numbers in the yellow interior hexagons must be $8,9,11,13,14,15$ .

Sum of each of 5 rows is 400/5=80. Following Magic Hexagon solution, 12 numbers from 17 to 33 are arranged in outer most ring. Arranging remaining 7 numbers from 5 to 15 is quite simple to get the solution. Center hexagon ends with number 5.

The sum of all the numbers is $400$ , so this divided by the $5$ horizontal rows gives a magical constant $M$ for each row of $M = \frac{400}{5} = 80$ .

The edge rows have $3$ numbers each, and the only way to make $3$ of the given numbers add up to $80$ is to use the numbers $15$ or greater ( $15$ , $17$ , $23$ , $24$ , $25$ , $26$ , $27$ , $28$ , $29$ , $30$ , $31$ , $32$ , and $33$ ). That's $13$ numbers for $12$ spots, and the sum of these $13$ numbers is $340$ . If $m$ is the number in the list that is not in one of the edge rows, then $15 \leq m \leq 33$ .

Let $E_V$ be the sum of the exterior cells at each vertex of the hexagon, and let $E_S$ be the sum of the exterior cells at the center of each side of the hexagon. Using $12$ of the $13$ numbers from above, $E_V + E_S = 340 - m$ , and since each of the $6$ edge rows add up to $M = 80$ , $2E_V + E_S = 6M = 480$ . These two equations solve to $E_V = m + 140$ .

Let $I_R$ be the sum of the ring of cells in the interior of the hexagon, and let $I_C$ be the number in the center of the hexagon. Since each of the $6$ rows with length $4$ add up to $M = 80$ , $2E_S + 2I_R = 6M = 480$ , or $E_S + I_R = 240$ . Since each of the $5$ horizontal rows add up to $M = 80$ , $E_V + E_S + I_R + I_C = 5M = 400$ . These two equations solve to $E_V + I_C = 160$ .

Combining $E_V = m + 140$ and $E_V + I_C = 160$ solves to $I_C = 20 - m$ . Since $m \geq 15$ , that means $I_C \leq 5$ , and since $5$ is the smallest number available, $I_C = \boxed{5}$ .

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For the rest of the hexagon, since $I_C = 5$ , then from $I_C = 20 - m$ we have $m = 15$ , and from $E_V = m + 140$ we have $E_V = 155$ . Disregarding similar rotational symmetries, there are only $3$ different ways the edge rows could be filled:

Filling in $5$ as the center number and $x$ as one of the other interior cells, and using the fact that each row adds up to $M = 80$ , we can deduce the following:

In the first case, there is an x and a 31 - x, but no two numbers 15 or less can add up to 31. In the second and third case, there is an x, 29 - x, and x - 1, and with the given numbers x = 14. However in the second case, this leads to 21 - x = 7, which is not an option. Therefore, it is the third case where x = 14, which solves to the following: