Henry Ernest Dudeney/Puzzles and Curious Problems/305 - Domino Frames/Solution
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Puzzles and Curious Problems by Henry Ernest Dudeney: $305$
- Domino Frames
- 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.
Solution
- 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.
Sources
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $305$. -- Domino Frames
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $486$. Domino Frames