DIGITAL DIVISION.
(
Money Puzzles)
It is another good puzzle so to arrange the nine digits (the nought
excluded) into two groups so that one group when divided by the other
produces a given number without remainder. For example, 1 3 4 5 8
divided by 6 7 2 9 gives 2. Can the reader find similar arrangements
producing 3, 4, 5, 6, 7, 8, and 9 respectively? Also, can he find the
pairs of smallest possible numbers in each case? Thus, 1 4 6 5 8 divided
by 7 3 2 9 is just as correct for 2 as the other example we have given,
but the numbers are higher.
Answer:
It is convenient to consider the digits as arranged to form fractions of
the respective values, one-half, one-third, one-fourth, one-fifth,
one-sixth, one-seventh, one-eighth, and one-ninth. I will first give the
eight answers, as follows:--
6729/13458 = 1/2
5823/17469 = 1/3
3942/15768 = 1/4
2697/13485 = 1/5
2943/17658 = 1/6
2394/16758 = 1/7
3187/25496 = 1/8
6381/57429 = 1/9
The sum of the numerator digits and the denominator digits will, of
course, always be 45, and the "digital root" is 9. Now, if we separate
the nine digits into any two groups, the sum of the two digital roots
will always be 9. In fact, the two digital roots must be either 9--9,
8--1, 7--2, 6--3, or 5--4. In the first case the actual sum is 18, but
then the digital root of this number is itself 9. The solutions in the
cases of one-third, one-fourth, one-sixth, one-seventh, and one-ninth
must be of the form 9--9; that is to say, the digital roots of both
numerator and denominator will be 9. In the cases of one-half and
one-fifth, however, the digital roots are 6--3, but of course the higher
root may occur either in the numerator or in the denominator; thus
2697/13485, 2769/13845, 2973/14865, 3729/18645, where, in the first two
arrangements, the roots of the numerator and denominator are
respectively 6--3, and in the last two 3--6. The most curious case of
all is, perhaps, one-eighth, for here the digital roots may be of any
one of the five forms given above.
The denominators of the fractions being regarded as the numerators
multiplied by 2, 3, 4, 5, 6, 7, 8, and 9 respectively, we must pay
attention to the "carryings over." In order to get five figures in the
product there will, of course, always be a carry-over after multiplying
the last figure to the left, and in every case higher than 4 we must
carry over at least three times. Consequently in cases from one-fifth to
one-ninth we cannot produce different solutions by a mere change of
position of pairs of figures, as, for example, we may with 5832/17496
and 5823/17469, where the 2/6 and 3/9 change places. It is true that the
same figures may often be differently arranged, as shown in the two
pairs of values for one-fifth that I have given in the last paragraph,
but here it will be found there is a general readjustment of figures and
not a simple changing of the positions of pairs. There are other little
points that would occur to every solver--such as that the figure 5
cannot ever appear to the extreme right of the numerator, as this would
result in our getting either a nought or a second 5 in the denominator.
Similarly 1 cannot ever appear in the same position, nor 6 in the
fraction one-sixth, nor an even figure in the fraction one-fifth, and so
on. The preliminary consideration of such points as I have touched upon
will not only prevent our wasting a lot of time in trying to produce
impossible forms, but will lead us more or less directly to the desired
solutions.
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