(related to Problem: Digital Division)
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: —
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 $\frac{2697}{13485}$, $\frac{2769}{13845}$, $\frac{2973}{14865}$, $\frac{3729}{18645}$, 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 $\frac{5832}{17496}$ and $\frac{5823}{17469}$, where the $\frac{2}{6}$ and $\frac{3}{9}$ 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 naught 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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