Tom Neill did a superb job working on this
Bernard's Bag problem. He really did explore a great number of
possibilities, and Tom is wondering if there are even more
investigations that he can make. Well done, Tom.
Here is a copy of the results that Tom came up with.
- The magic constant is 34.
There are 8 2x2 squares - each corner and each middle part of each
side makes a square.
- If you add 2 to each number in the square then the magic
constant becomes: +95-14+17x0-20+62 = 42
| 17 |
12 |
5 |
8 |
| 6 |
7 |
18 |
11 |
| 16 |
13 |
4 |
9 |
| 3 |
10 |
15 |
14 |
- If you double each number then the magic constant doubles to
equal 68.
- Or, you can make as square in which the magic constant is 17 by
halving each number:
| 7 1/2 |
5 |
1 1/2 |
3 |
| 2 |
2 1/2 |
8 |
4 4/8 |
| 7 |
5 64/128 |
1 |
3 256/512 |
| 16384/32768 |
4 |
6 1024/2048 |
6 |
- To make a square in which the magic constant is 38 you add one
to each number:
| 16 |
11 |
4 |
7 |
| 5 |
6 |
17 |
10 |
| 15 |
12 |
3 |
8 |
| 2 |
9 |
14 |
13 |
- You can also make a magic square that has a magic constant of
50:
| 19 |
14 |
7 |
10 |
| 8 |
9 |
20 |
13 |
| 18 |
15 |
6 |
11 |
| 5 |
12 |
17 |
16 |
- And, you can even make a magic square a magic constant of zero!
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
- Along the diagonals the squares can have the same totals along
the diagonals and still work - here is an example with the constant
of 50
| 19 |
8 |
18 |
5 |
| 14 |
9 |
15 |
12 |
| 7 |
20 |
6 |
17 |
| 10 |
13 |
11 |
16 |
However,
- You can't make a square with a magic constant of 32 - although
you could if you allowed fractions to be used!