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

Now, if we took every possible combination and tested it by multiplication, we should need to make no fewer than 30,240 trials, or, if we at once rejected the number 1 as a multiplier, 28,560 trials—­a task that I think most people would be inclined to shirk.  But let us consider whether there be no shorter way of getting at the results required.  I have already explained that if you add together the digits of any number and then, as often as necessary, add the digits of the result, you must ultimately get a number composed of one figure.  This last number I call the “digital root.”  It is necessary in every solution of our problem that the root of the sum of the digital roots of our multipliers shall be the same as the root of their product.  There are only four ways in which this can happen:  when the digital roots of the multipliers are 3 and 6, or 9 and 9, or 2 and 2, or 5 and 8.  I have divided the twenty-two answers above into these four classes.  It is thus evident that the digital root of any product in the first two classes must be 9, and in the second two classes 4.

Owing to the fact that no number of five figures can have a digital sum less than 15 or more than 35, we find that the figures of our product must sum to either 18 or 27 to produce the root 9, and to either 22 or 31 to produce the root 4.  There are 3 ways of selecting five different figures that add up to 18, there are 11 ways of selecting five figures that add up to 27, there are 9 ways of selecting five figures that add up to 22, and 5 ways of selecting five figures that add up to 31.  There are, therefore, 28 different groups, and no more, from any one of which a product may be formed.