Answer :
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 twofigure
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 singlefigure number.
This I call the
"digital root."
Thus, the digital root of 521 is 8, and of 697 it is 4.
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 twofigure 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
(thirtysix 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
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 3×6 and 8×6 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 × 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 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.
