# Solution

(related to Problem: More Mixed Fractions)

The point of the present puzzle lies in the fact that the numbers $15$ and $18$ are not capable of solution. There is no way of determining this without trial. Here are answers for the ten possible numbers: —

• $9\frac{5472}{1368} = 13;$
• $9\frac{6435}{1287} = 14;$
• $12\frac{3576}{894} = 16;$
• $6\frac{13258}{947} = 20;$
• $15\frac{9432}{786} = 27;$
• $24\frac{9756}{813} = 36;$
• $27\frac{5148}{396} = 40;$
• $65\frac{1892}{473} = 69;$
• $59\frac{3614}{278} = 72;$
• $75\frac{3648}{192} = 94.$

I have only found the one arrangement for each of the numbers $16, 20, and 27;$ but the other numbers are all capable of being solved in more than one way. As for $15$ and $18,$ though these may be easily solved as a simple fraction, yet a "mixed fraction" assumes the presence of a whole number; and though my own idea for dodging the conditions is the following, where the fraction is both complex and mixed, it will be fairer to keep exactly to the form indicated:—

• $3\frac{8952}{746}/1 = 15$
• $9\frac{5742}{638}/1 = 18$

I have proved the possibility of solution for all numbers up to $100,$ except $1, 2, 3, 4, 15,$ and $18.$ The first three are easily shown to be impossible. I have also noticed that numbers whose digital root is $8$ — such as $26, 35, 44, 53,$ etc.—seem to lend themselves to the greatest number of answers. For the number $26$ alone I have recorded no fewer than twenty-five different arrangements, and I have no doubt that there are many more.

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### References

#### Project Gutenberg

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

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