This puzzle is both easy and difficult, for it is a very simple matter to
find one of the multipliers, which is 86. If we multiply 8 by 86, all we
need do is to place the 6 in front and the 8 behind in order to get the
correct answer, 688. But the second number is not to be found by mere
trial. It is 71, and the number to be multiplied is no less than
1639344262295081967213114754098360655737704918032787. If you want to
multiply this by 71, all you have to do is to place another 1 at the
beginning and another 7 at the end--a considerable saving of labour!
These two, and the example shown by the wizard, are the only two-figure
multipliers, but the number to be multiplied may always be increased.
Thus, if you prefix to 41096 the number 41095890, repeated any number of
times, the result may always be multiplied by 83 in the wizard's peculiar
manner.
If we add the figures of any number together and then, if necessary,
again add, we at last get a single-figure number. This I call the
"digital root." Thus, the digital root of 521 is 8, and of 697 it is 4.
This digital analysis is extensively dealt with in _A. in M._ Now, it is
evident that the digital roots of the two numbers required by the puzzle
must produce the same root in sum and product. This can only happen when
the roots of the two numbers are 2 and 2, or 9 and 9, or 3 and 6, or 5
and 8. Therefore the two-figure multiplier must have a digital root of 2,
3, 5, 6, 8, or 9. There are ten such numbers in each case. I write out
all the sixty, then I strike out all those numbers where the second
figure is higher than the first, and where the two figures are alike
(thirty-six numbers in all); also all remaining numbers where the first
figure is odd and the second figure even (seven numbers); also all
multiples of 5 (three more numbers). The numbers 21 and 62 I reject on
inspection, for reasons that I will not enter into. I then have left, out
of the original sixty, only the following twelve numbers: 83, 63, 81, 84,
93, 42, 51, 87, 41, 86, 53, and 71. These are the only possible
multipliers that I have really to examine.
My process is now as curious as it is simple in working. First trying 83,
I deduct 10 and call it 73. Adding 0's to the second figure, I say if
30000, etc., ever has a remainder 43 when divided by 73, the dividend
will be the required multiplier for 83. I get the 43 in this way. The
only multiplier of 3 that produces an 8 in the digits place is 6. I
therefore multiply 73 by 6 and get 438, or 43 after rejecting the 8. Now,
300,000 divided by 73 leaves the remainder 43, and the dividend is 4,109.
To this 1 add the 6 mentioned above and get 41,096 x 83, the example
given on page 129.
In trying the even numbers there are two cases to be considered. Thus,
taking 86, we may say that if 60000, etc., when divided by 76 leaves
either 22 or 60 (because 3x6 and 8x6 both produce 8), we get a solution.
But I reject the former on inspection, and see that 60 divided by 76 is
0, leaving a remainder 60. Therefore 8 x 86 = 688, the other example. It
will be found in the case of 71 that 100000, etc., divided by 61 gives a
remainder 42, (7 x 61 = 427) after producing the long dividend at the
beginning of this article, with the 7 added.
The other multipliers fail to produce a solution, so 83, 86, and 71 are
the only three possible multipliers. Those who are familiar with the
principle of recurring decimals (as somewhat explained in my next note on
No. 83, "The Ribbon Problem") will understand the conditions under which
the remainders repeat themselves after certain periods, and will only
find it necessary in two or three cases to make any lengthy divisions. It
clearly follows that there is an unlimited number of multiplicands for
each multiplier.