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.