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
(a,b,c)
=
(4,2,1)
®
562748139
(2,4,1)
®
X
(1,4,2)
®
X.
For a+b+c=11, T=14
(a,b,c) =
(8,2,1)® 392485167
(7,3,1)® 293476158
(6,3,2)® X, (6,4,1)® X, (5,4,2)® X
(4,6,1)® X, (4,5,2)® X, (3,7,1)® X
(3,6,2)® 176539248
(2,8,1)® X, (2,6,3)® X
(2,5,4)® 365728419
(1,8,2)® X (1,7,3)® X
(1,6,4)® 356719428.
There are 6 solutions with T < 15, and therefore 6 corresponding
ones with T > 15, for a total of twelve. All are listed below: