In this case a certain amount of mere “trial” is unavoidable.  But there are two kinds of “trials”—­those that are purely haphazard, and those that are methodical.  The true puzzle lover is never satisfied with mere haphazard trials.  The reader will find that by just reversing the figures in 23 and 46 (making the multipliers 32 and 64) both products will be 5,056.  This is an improvement, but it is not the correct answer.  We can get as large a product as 5,568 if we multiply 174 by 32 and 96 by 58, but this solution is not to be found without the exercise of some judgment and patience.

82.—­THE TEN COUNTERS.

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 x 2 = 6,970, and 6,970 x 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 x 64 = 58,560, and 732 x 80 = 58,560.

83.—­DIGITAL MULTIPLICATION.

The solution that gives the smallest possible sum of digits in the common product is 23 x 174 = 58 x 69 = 4,002, and the solution that gives the largest possible sum of digits, 9x654 =18x327=5,886.  In the first case the digits sum to 6 and in the second case to 27.  There is no way of obtaining the solution but by actual trial.

84.—­THE PIERROT’S PUZZLE.

There are just six different solutions to this puzzle, as follows:—­

8 multiplied by 473 equals 3784
9      "        351   "    3159
15      "         93   "    1395
21      "         87   "    1287
27      "         81   "    2187
35      "         41   "    1435

It will be seen that in every case the two multipliers contain exactly the same figures as the product.