|
In the figure, place the numbers 1-13 at the locations a-m such that
each of the six sides and each of the six lines through the center has the same sum
(all 12 sums equal.) To aid in comparisons, when submitting answers, minimize 'a', then 'b'. Solve for '1' both at 'a' and 'b'. |
a b c l d k m e j f i h g |
There are 4 solutions:
8 1 12 9 1 11
3 5 2 6
10 7 4 10 7 4
9 11 8 12
2 13 6 3 13 5
1 8 12 1 9 11
11 4 12 4
9 7 5 8 7 6
10 3 10 2
2 6 13 3 5 13
Claudio Baiocchi's solution is below:
The solution I join is not the shorter one: one could directly use the
properties (where S denotes the common value of the 12 sums):
S = 3 M ; 6 S = 5 M + 91
proven later on. However I think is interesting to firstly explore all the
consequences of "constant sums along the 6 lines through the center"; and to
impose the sums along the sides only in a second stage.
-------------------------------------
PART 1. LINES THROUGH THE CENTER.
There are 3 families of obvious solutions:
1.1 We choose M=13; then we couple 1 with 12, 2 with 11 and so on up to
6 with 7; thus obtaining 6 couples with sum 13. Putting these couples, in
any order, in points opposite with respect to the center we get solutions
with constant sum 26.
1.2 We choose M=1; then we couple 2 with 13, 3 with 12 and so on, up to
7 with 8; thus obtaining 6 couples with sum 15. Putting these couples, in
any order, in points opposite with respect to the center we get solutions
with constant sum 16; remark that these solutions are the complement (any
number x is replaced by 14-x) of the previous ones.
1.3 We choose M=7; then we couple 1 with 13, 2 with 12 and so on up to 6
with 8; thus obtaining 6 couples with sum 14. Putting these couples, in any
order, in points opposite with respect to the center we get solutions with
constant sum 21.
Let us proove that there are no further solution. In fact, let S be the
common value of the 6 sums through the center; adding together these sums we
get:
6 S = A + B + ... + L + 6 M = 5 M + 91
and evaluating this equation modulo 6 we get that the reminder of M/6 is 1;
thus M can only take the values 1, 7 or 13; accordingly our formula gives
for S the values 26, 21 and 16 respectively.
PART 2. SUMS ALONG THE SIDES
Let us impose that also two opposite sides (e.g. "A-B-C" and "G-H-I") give
raise to the sum S. Then:
S = G + H + I = (S - M - A)+(S - M - B)+
(S - M - C) = 3 S - 3 M - (A + B + C)
= 2 S - 3 M
thus S = 3 M. This excludes the case M=1, S=26 and the case M=13, S=16;
leaving the unique possibility M=7, S=21.
PART 3. THE TRUE PROBLEM.
Let us distinguish two cases:
Case A. The value 1 is not in a vertex. After a rotation we can assume it is
in B as asked; possibly after a symmetry along the line "B-M-H" we can
assume A less than C, as required. From A+B+C=21, B=1, A less than C, and
the value 7 being already used for M, we are left with only two
possibilities:
A=8, C=12; or A=9, C=11. Both these schemes can be completed in a unique
way:
8 1 12 9 1 11
3 5 2 6
10 7 4 10 7 4
9 11 8 12
2 13 6 3 13 5
Case B. The value 1 is in a vertex. After a rotation we can assume it is in
A, as asked; possibly after a symmetry along the line "A-M-F" we can assume
B less than L, as required. We get again two solutions, say: 1 8 12 1 9 11
11 4 12 4
9 7 5 8 7 6
10 3 10 2
2 6 13 3 5 13
In fact for B, C, K, L we can only use the values 8, 9, 11, 12; furthermore,
because we want B less than L, the value 12 is forbidden for B and the value