Copyright © University of Cambridge. All rights reserved.
Thank you for your solutions to John from State College Area High
School, Pennsylvania, USA, Andrei Lazanu from School No. 205,
Bucharest, Romania, Sarah from Madras College St Andrew's Scotland
and Patrick and his friend David Lee Yick Ming from Hkma David Li
Kwok Po College, Hong Kong. This is John's solution.
First, I show that for every solution with a magic total $T$ there
is a corresponding solution with magic total $30-T$. Consider what
would happen if every element $x$ in a solution were replaced with
the element $10-x$. All elements from $0$-$9$ would still be used,
and all of the rows would have a sum $30$ minus the original total.
Since all the row-sums were initially equal, they would still be,
but each sum would be $30$ minus the old one, that is $30-T$.
Now I'll find all possible magic arrangements. First, I'll
introduce my notation. Call the top centre vertex (occupied by the
$2$ in the example picture) $a$. Call the lower left one $b$
(currently $5$). Call the lower right one ($4$) $c$. Arbitrarily,
I'll set $b> c$, since solutions that are reflections of each
other are considered to be identical.
Now, it is clear that the sum of the totals of all four rows will
be equal to $(1+2+3+4+5+6+7+8+9)+a+b+c$, since $a,\ b$, and $c$ are
the only elements to appear in all four rows. This simplifies to
$45+a+b+c$. If all four rows are equal, the magic total, in terms
of $a,\ b$, and $c$, will be $(45+a+b+c)/4$ since there are four
rows. For the magic sum to be integral, $(a+b+c)$ must be one less
than a multiple of 4. Possible sums for $a+b+c$ are then $3$
($T=12$), $7$ ($T=13$), $11$ ($T=14$), and $15$ ($T=15$). We need
not consider any arrangements with $T> 15$, as they would have
corresponding arrangements with $T< 15 $as discussed in the
first paragraph. We may also ignore $a+b+c=3$ as values that small
cannot be chosen from the integers $1$ through $9$. The possible
sums $a+b+c$ are therefore 7, 11, and 15, with $T=13,\ 14$, and
$15$ respectively.
We may further narrow the search by eliminating the case
$a+b+c=15$. When $a+b+c=15$, the magic sum $T$ also =15. Since $a$
and $b$ appear together in one row, the other number needed to give
a sum of 15 in that row is $c$. However $c$ cannot appear in that
row as it must appear elsewhere. Therefore $a+b+c=7$ or $11$. Now
I'll use a brute force search to find all solutions. All possible
arrangements are listed below. I checked each of them to determine
if it yielded a Magic W. Those that did have the numbers listed as
they appear from left to right in the W. Those that did not are
marked only with an 'X'.
For $a+b+c=7,\ T=13$
\begin{eqnarray} % the * suppresses equation
numbering (a,b,c)&= \\ % & is the column separator
(4,2,1)&\rightarrow &562748139 \\ (2,4,1)&\rightarrow
&X \\ (1,4,2)&\rightarrow &X.\\ \end{eqnarray}
For $a+b+c=11,\ T=14$
\begin{eqnarray} (a,b,c)= \\ (8,2,1)\rightarrow
392485167 \\ (7,3,1)\rightarrow 293476158 \\ (6,3,2)\rightarrow X,\
(6,4,1)\rightarrow X,\ (5,4,2)\rightarrow X \\ (4,6,1)\rightarrow
X,\ (4,5,2)\rightarrow X,\ (3,7,1)\rightarrow X \\
(3,6,2)\rightarrow 176539248 \\ (2,8,1)\rightarrow X,\
(2,6,3)\rightarrow X \\ (2,5,4)\rightarrow 365728419 \\
(1,8,2)\rightarrow X \ (1,7,3)\rightarrow X \\ (1,6,4)\rightarrow
356719428. \\ \end{eqnarray}
There are $6$ solutions with $T< 15$, and therefore $6$
corresponding ones with $T> 15$, for a total of twelve. All are
listed below:
$T=13$ $562748139$
$T=14$ $392485167$, $293476158$, $176539248$, $365728419$,
$356719428$
$T=16$ $718625943$, $817634952$, $934571862$, $745382691$,
$754391682$
$T=17$ $548362971$