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. 85.--THE CAB NUMBERS. The highest product is, I think, obtained by multiplying 8,745,231 by 96--namely, 839,542,176. Dealing here with the problem generally, I have shown in the last puzzle that with three digits there are only two possible solutions, and with four digits only six different solutions. These cases have all been given. With five digits there are just twenty-two solutions, as follows:-- 3 x 4128 = 12384 3 x 4281 = 12843 3 x 7125 = 21375 3 x 7251 = 21753 2541 x 6 = 15246 651 x 24 = 15624 678 x 42 = 28476 246 x 51 = 12546 57 x 834 = 47538 75 x 231 = 17325 624 x 78 = 48672 435 x 87 = 37845 ------ 9 x 7461 = 67149 72 x 936 = 67392 ------ 2 x 8714 = 17428 2 x 8741 = 17482 65 x 281 = 18265 65 x 983 = 63985 ------ 4973 x 8 = 39784 6521 x 8 = 52168 14 x 926 = 12964 86 x 251 = 21586