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 he showed Hawkhurst the solution privately'

See answer