Place the numbers 1 to 9 into the 3x3 square so that:
(a) the sum of the three numbers in the top row is 11;
(b) the sum of the three numbers in the bottom row is 18;
(c) the sum of the three numbers in the left column is 21; and
(d) the sum of the three numbers in the right column is 17;
Please send in your answers using the letters below.
Use the format:
A =
B =
etc.
Solution to the Problem:
B = 2
C = 3
D = 7
E = 4
F = 5
G = 8
H = 1
I = 9
We are given the following:
A + B + C = 11
G + H + I = 18
C + F + I = 17
A + D + G = 21
From the clues above, observe that:
There are only five possibilities for A, B, and C:
(6,3,2), (6,4,1), (5,4,2), (8,2,1), and (7,3,1)
There are only three possibilities for A, D, and G:
(9,8,4), (8,7,6), and (9,7,5)
Therefore, A cannot be 1, 2, 3, or 9 since they are not in the
intersections of the two sets above.
I tried A = 4, but the numbers did not work.
Then I tried A = 5 and again the numbers did not work.
When I tried A = 6, I found a solution.