As I pointed out, it is quite easy so to arrange the counters that they
shall form a pair of simple multiplication sums, each of which will give
the same product--in fact, this can be done by anybody in five minutes
with a little patience. But it is quite another matter to find that pair
which gives the largest product and that which gives the smallest
product.
Now, in order to get the smallest product, it is necessary to select as
multipliers the two smallest possible numbers. If, therefore, we place 1
and 2 as multipliers, all we have to do is to arrange the remaining
eight counters in such a way that they shall form two numbers, one of
which is just double the other; and in doing this we must, of course,
try to make the smaller number as low as possible. Of course the lowest
number we could get would be 3,045; but this will not work, neither will
3,405, 3,45O, etc., and it may be ascertained that 3,485 is the lowest
possible. One of the required answers is 3,485 × 2 = 6,970, and 6,970 ×
1 = 6,970.
The other part of the puzzle (finding the pair with the highest product)
is, however, the real knotty point, for it is not at all easy to
discover whether we should let the multiplier consist of one or of two
figures, though it is clear that we must keep, so far as we can, the
largest figures to the left in both multiplier and multiplicand. It will
be seen that by the following arrangement so high a number as 58,560 may
be obtained. Thus, 915 × 64 = 58,560, and 732 × 80 = 58,560.