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 Mathematiques" 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 x
483 = 5,796. I now also propose to include the cases where there are
one, four, and four digits, such as 4 x 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,
5, 6, 7, 8 and 9. I arranged them on the table in two groups, as shown
in the illustration, so as to form two multiplication sums, and found
that both sums gave the same product. You will find that 158 multiplied
by 23 is 3,634, and that 79 multiplied by 46 is also 3,634. Now, the
puzzle I propose is to rearrange the counters so as to get as large a
product as possible. What is the best way of placing them? Remember both
groups must multiply to the same amount, and there must be three
counters multiplied by two in one case, and two multiplied by two
counters in the other, just as at present.

82.--THE TEN COUNTERS.

In this case we use the nought in addition to the 1, 2, 3, 4, 5, 6, 7,
8, 9. The puzzle is, as in the last case, so to arrange the ten counters
that the products of the two multiplications shall be the same, and you
may here have one or more figures in the multiplier, as you choose. The
above is a very easy feat; but it is also required to find the two
arrangements giving pairs of the highest and lowest products possible.
Of course every counter must be used, and the cipher may not be placed
to the left of a row of figures where it would have no effect. Vulgar
fractions or decimals are not allowed.

83.--DIGITAL MULTIPLICATION.

Here is another entertaining problem with the nine digits, the nought
being excluded. Using each figure once, and only once, we can form two
multiplic