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.—solution]
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.