Henry Ernest Dudeney/Puzzles and Curious Problems/305 - Domino Frames/Solution

Puzzles and Curious Problems by Henry Ernest Dudeney: $305$

Take an ordinary set of $28$ dominoes and return double $3$, double $4$, double $5$, and double $6$ to the box as not wanted.

Now, with the remainder form three square frames, in the manner shown, so that the pips in every side shall add up alike.

In the example given the sides sum to $15$.

If this were to stand, the sides of the other two frames must also sum to $15$.

But you can take any number you like, and it will be seen that it is not required to place $6$ against $6$, $5$ against $5$, and so on, as in play.

The three diagrams show a solution.

The sum of all the pips is $132$.

One-third of this is $44$.

First divide the dominoes into any three groups of $44$ pips each.

Then, if we decide to try $12$ for the sum of the sides, $4$ times $12$ being $4$ more than $44$,

we must arrange in every case that the four corners in a frame shall sum to $4$.

The rest is done by trial and exchanges from one group to another of dominoes containing an equal number of pips.