they formed a simple addition sum, the two upper rows of figures
producing the sum in the lowest row. But the most surprising point was
this: that he had so arranged them that the addition in A gave the
smallest possible sum, that the addition in C gave the largest possible
sum, and that all the nine digits in the three totals were different.
The puzzle is to show how this could be done. No decimals are allowed
and the nought may not appear in the hundreds place.


80.--THE THREE GROUPS.

There appeared in "Nouvelles Annales de Mathématiques" the following
puzzle as a modification of one of my "Canterbury Puzzles." Arrange the
nine digits in three groups of two, three, and four digits, so that the
first two numbers when multiplied together make the third. Thus, 12 ×
483 = 5,796. I now also propose to include the cases where there are
one, four, and four digits, such as 4 × 1,738 = 6,952. Can you find all
the possible solutions in both cases?


81.--THE NINE COUNTERS.

[Illustration:

(1)(5)(8) (7)(9)
(2)(3) (4)(6)

]

I have nine counters, each bearing one of the nine digits, 1, 2, 3, 4,