Henry Ernest Dudeney/Modern Puzzles/178 - The Seven-Pointed Star/Solution

by : $178$

 * The Seven-Pointed Star

Solution

 * Dudeney-Modern-Puzzles-178-solution.png

Place $5$ in the top point.

Let the $4$ numbers in the horizontal line, that is $7$, $11$, $9$, $3$, be such that:
 * the two outside numbers sum to $10$
 * the inner numbers sum to $20$
 * the difference between the two outer numbers is twice the difference between the two inner numbers.

Then their complements with $15$ are placed in the relative positions indicated by the dotted lines.

The remaining $4$ numbers, that is $13$, $2$, $14$, $1$ are easily adjusted.

From this arrangement we can get $3$ others.


 * $(1): \quad$ Change the $13$ with the $1$ and the $14$ with the $2$.


 * $(2)$ and $(3): \quad$ Substitute in each of those two above arrangements its difference from $15$.

There are $56$ different arrangements, counting complements.

Class $\text I$ is those as above, where pairs in the positions $7 - 8$, $13 - 2$, $3 - 12$, $14 - 1$ all add to $15$, and there are $20$ such cases.

Class $\text {II}$ includes cases where pairs in the positions $7 - 2$, $8 - 13$, $3 - 1$, $12 - 14$ all add to $15$, and there are $20$ such cases.

Class $\text {iII}$ includes cases where pairs in the positions $7 - 8$, $13 - 2$, $3 - 1$, $12 - 14$ all add to $15$, and there are $16$ such cases.