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