# Solution

(related to Problem: The Card Frame Puzzle)

The sum of all the pips on the ten cards is $55.$ Suppose we are trying to get $14$ pips on every side. Then $4$ times $14$ is $56.$ But each of the four corner cards is added in twice so that $55$ deducted from $56,$ or $1,$ must represent the sum of the four corner cards. This is clearly impossible; therefore $14$ is also impossible. But suppose we came to trying $18.$ Then $4$ times $18$ is $72,$ and if we deduct $55$ we get $17$ as the sum of the corners. We need then only try different arrangements with the four corners always summing to $17,$ and we soon discover the following solution:—

The final trials are very limited in number, and must with a little judgment either bring us to a correct solution or satisfy us that a solution is impossible under the conditions we are attempting. The two center cards on the upright sides can, of course, always be interchanged, but I do not call these different solutions. If you reflect in a mirror you get another arrangement, which also is not considered different. In the answer given, however, we may exchange the $5$ with the $8$ and the $4$ with the $1.$ This is a different solution. There are two solutions with $18,$ four with $19,$ two with $20,$ and two with $22$ — ten arrangements in all. Readers may like to find all these for themselves.

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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