# New Scientist Enigma 587 – Find the Fractions

*by John Magill*

#### From New Scientist #1740, 27th October 1990

\[\Large\begin{array} {|c|c|c|}

\hline \frac{29}{e} & \frac{14}{c} & \frac{11}{e} \\

\hline\frac{7}{d} & \frac{8}{c} & \frac{5}{a} \\

\hline\frac{53}{e} & \frac{2}{c} & \frac{7}{b} \\

\hline \end{array}\]

In the figure each letter represents a number, the same letter standing for the same number wherever it appears. The nine entries are vulgar fractions in reduced form, that is, in each fraction the numerator and denominator have no common factor apart from 1. When completed, the figure is a magic square where each of the rows, columns and diagonals sum to the same value, a value greater than a half and less than 1.

What is the magic square revealed by substituting, for each letter, its number?

In an additive magic square the rows, columns and diagonals all have the same sum, which is equal to three times the centre value.

The first row has only the variables \(e\) and \(c\) so we can establish an equation to relate these two variables by using the row sum:

\[\frac{29}{e}+\frac{14}{c}+\frac{11}{e}=\frac{24}{c} \hspace{10pt}\Rightarrow \hspace{10pt}\frac{40}{e}=\frac{10}{c} \hspace{10pt}\Rightarrow \hspace{10pt}e=4c \tag{1}\]

Now consider the first column sum and use equation (1) to substitute for \(e\) in terms of \(c\):

\[\frac{29}{e}+\frac{7}{d}+\frac{53}{e}=\frac{24}{c} \hspace{10pt}\Rightarrow \hspace{10pt}\frac{7}{d}=\frac{(96-29-53)}{4c}=\frac{14}{4c} \hspace{10pt}\Rightarrow \hspace{10pt}d=2c \tag{2}\]

Similarly the bottom row and right hand column respectively yield \(b\) and \(a\) in terms of \(c\):

\[\frac{53}{e}+\frac{2}{c}+\frac{7}{b}=\frac{24}{c} \hspace{10pt}\Rightarrow \hspace{10pt}\frac{7}{b}=\frac{(96-8-53)}{4c}=\frac{35}{4c} \hspace{10pt}\Rightarrow \hspace{10pt}b=\frac{4c}{5}\tag{3}\]\[\frac{11}{e}+\frac{5}{a}+\frac{7}{b}=\frac{24}{c} \hspace{10pt}\Rightarrow \hspace{10pt}\frac{5}{a}=\frac{(96-11-35)}{4c}=\frac{50}{4c} \hspace{10pt}\Rightarrow \hspace{10pt}a=\frac{2c}{5}\tag{4}\]

Since the values \(a\) .. \(e\) are all integer, we can see that \(c\) must be a multiple of 5. Letting \(c=5k\) we then have:

\[a=2k,\hspace{10pt}b=4k,\hspace{10pt}c=5k,\hspace{10pt}d=10k,\hspace{10pt}e=20k\tag{5}\]

We are told that the row/column/diagonal sum satisfies:

\[\frac{1}{2} < \frac{24}{c} < 1\tag{6}\]

which shows that \(24<5k<48\) and hence sets the range of possible \(k\) values as \(5 ... 9\). But, since all the fractions in the magic square are in their lowest terms, \(k\) cannot be 5, 6, 7 or 8, leaving \(k=9\) as the only solution, giving:\[a=18,\hspace{10pt}b=36,\hspace{10pt}c=45,\hspace{10pt}d=90,\hspace{10pt}e=180\tag{7}\]and the magic square:

\[\Large\begin{array} {|c|c|c|}

\hline \frac{29}{180} & \frac{14}{45} & \frac{11}{180} \\

\hline\frac{7}{90} & \frac{8}{45} & \frac{5}{18} \\

\hline\frac{53}{180} & \frac{2}{45} & \frac{7}{36} \\

\hline \end{array}\]