The Coinage Puzzle
(
THE PROFESSOR'S PUZZLES)
He made a rough diagram, and placed a crown and a florin in two of the divisions, as indicated in the illustration.
"Now," he continued, "place the fewest possible current English coins in the seven empty divisions, so that each of the three columns, three rows, and two diagonals shall add up fifteen shillings. Of course, no division may be without at least one coin, and no two divisions may contain the same value."
"But how can the coins affect the question?" asked Grigsby.
"That you will find out when you approach the solution."
"I shall do it with numbers first," said Hawkhurst, "and then substitute coins."
Five minutes later, however, he exclaimed, "Hang it all! I can't help getting the 2 in a corner. May the florin be moved from its present position?"
"Certainly not."
"Then I give it up."
But Grigsby and I decided that we would work at it another time, so the Professor showed Hawkhurst the solution privately, and then went on with his chat.
Answer:
The point of this puzzle turns on the fact that if the magic square were to be composed of whole numbers adding up 15 in all ways, the two must be placed in one of the corners. Otherwise fractions must be used, and these are supplied in the puzzle by the employment of sixpences and half-crowns. I give the arrangement requiring the fewest possible current English coins—fifteen. It will be seen that the amount in each corner is a fractional one, the sum required in the total being a whole number of shillings.
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