Page:Amusements in mathematics.djvu/170

158 87.—THE NUMBER CHECKS PUZZLE.

the ten checks into the following three groups: 7 1 5—4 6—3 2 8 9 0, and the first multiplied by the second produces the third.

88.—DIGITAL DIVISION.

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 oneninth 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.

89.—ADDING THE DIGITS.

smallest possible sum of money is £1, 8s. $9 3⁄4$d., the digits of which add to 25.

90.—THE CENTURY PUZZLE.

problem of expressing the number 100 as a mixed number or fraction, using all the nine digits once, and once only, has, like all these digital puzzles, a fascinating side to it. The merest tyro can by patient trial obtain correct results, and there is a singular pleasure in discovering and recording each new arrangement akin to the delight of the botanist in finding some long-sought plant. It is simply a matter of arranging those nine figures correctly, and yet with the thousands of possible combinations that confront us the task is not so easy as might at first appear, if we are to get a considerable number of results. Here are eleven answers, including the one I gave as a specimen:—

$96 2148⁄537$,$96 1752⁄438$,$96 1428⁄357$,$94 1578⁄263$,$91 7542⁄836$,

$91 5823⁄647$,$91 5742⁄638$,$82 3546⁄197$,$81 7524⁄296$,

$81 5643⁄297$,$3 69258⁄714$

Now, as all the fractions necessarily represent whole numbers, it will be convenient to deal with them in the following form: 96 + 4, 94 + 6, 91 + 9, 82 + 18, 81 + 19, and 3 + 97.

With any whole number the digital roots of the fraction that brings it up to 100 will always be of one particular form. Thus, in the case of 96 + 4, one can say at once that if any answers are obtainable, then the roots of both the numerator and the denominator of the fraction will be 6. Examine the first three arrangements given above, and you will find that this is so. In the case of 94 + 6 the roots of the numerator and denominator will be respectively 3—2, in the case of 91 + 9 and of 82 + 18 they will be 9—8, in the case of 81 + 19 they will be 9—9, and in the case of 3+97 they will be 3—3. Every fraction that can be employed has, therefore, its particular digital root form, and you are only wasting your time in unconsciously attempting to break through this law.

Every reader will have perceived that certain whole numbers are evidently impossible. Thus, if there is a 5 in the whole number, there will also be a nought or a second 5 in the fraction, which are barred by the conditions. Then multiples of 10, such as 90 and 80, cannot of course occur, nor can the whole number conclude with a 9, like 89 and 79, because the fraction, equal to 11 or 21, will have 1 in the last place, and will therefore repeat a figure. Whole numbers that repeat a figure, such as 88 and 77, are also clearly useless. These cases, as I have said, are all obvious to every reader. But when I declare