(related to Problem: The Century Puzzle)

The 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: —

- $96\frac {2148}{537}$
- $96\frac {1752}{438}$
- $96\frac {1428}{357}$
- $94\frac {1578}{263}$
- $91\frac{ 7524}{836}$
- $91\frac {5823}{647}$
- $91\frac {5742}{638}$
- $82\frac {3546}{197}$
- $81\frac {7524}{396}$
- $81\frac {5643}{297}$
- $3\frac {69258}{714}$

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 naught or a second $5$ in the fraction, which is 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 that such combinations as $98 + 2,$ $92 + 8$, $86 + 14,$ $83 + 17,$ $74 + 26,$ etc., etc., are to be at once dismissed as impossible, the reason is not so evident, and I unfortunately cannot spare space to explain it.

But when all those combinations have been struck out that are known to be impossible, it does not follow that all the remaining "possible forms" will actually work. The elemental form may be right enough, but there are other and deeper considerations that creep in to defeat our attempts. For example, $98 + 2$ is an impossible combination, because we are able to say at once that there is no possible form for the digital roots of the fraction equal to $2.$ But in the case of $97 + 3$ there is a possible form for the digital roots of the fraction, namely, $6—5,$ and it is only on further investigation that we are able to determine that this form cannot in practice be obtained, owing to curious considerations. The working is greatly simplified by a process of elimination, based on such considerations as that certain multiplications produce a repetition of figures, and that the whole number cannot be from $12$ to $23$ inclusive since in every such case sufficiently small denominators are not available for forming the fractional part.

**Dudeney, H. E.**: "Amusements in Mathematics", The Authors' Club, 1917

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