[Guest]

Substitute each of the capital letters in this 3x3 square by a different digit from 1 to 9 such that the sum of digits in each of the four 2x2 subsquares is equal to 9*E.

A   B   C

D   E   F

G   H   I 

What will be the arrangement, if keeping all the other conditions unaltered, the sum of digits in each of the four 2x2 subsquares is equal to 7*E?

Note: The reflections and rotations of a given arrangement are treated as the same solution.

[Pickett]

PART 1

So, we know that E must be either 1, 2, or 3...because any higher digit and there would be no way that 4 single digits could add up to 9*E.

If E = 1...we need A + B + D + E = 9...or A + B + D = 8... where A, B, and D cannot be 1 (since E is 1)...and there are no possible combinations that work...so E does not equal 1.

Let's move to E = 3...then we need A + B + D + E = 27...or A + B + D = 24...and there is only one possibility for that 7 + 8 + 9...but that would only give us one of the 4 2x2 squares...so we know that E must be 2.

E = 2

A + B + D + E = 18

A + B + D = 16

the combinations that work for this equation are:

1 7 8

1 6 9

3 4 9

3 6 7

4 5 7

So, What we need to find out of these 5 combinations are 4 of them such that 4 of the numbers are used only once, and 4 of them are used twice...

if we remove the 3 6 7 combination, we can then see that 1, 4, 7, and 9 are used twice...3, 5, 6, and 8 are used only once...so, now you just have to arrange accordingly...and our answer is

8 1 6

7 2 9

5 4 3

PART 2

We can then use the same sort of logic to determine that if they add to 7*E...E must be 3...and we get combinations of

A + B + D = 18

1 8 9

2 7 9

4 5 9

4 6 8

5 6 7

Which if we remove the 4 5 9, we see that 1 2 4 5 are used once, 6 7 8 9 are used twice...so our answer is

1 8 4

9 3 6

2 7 5

Edited July 28, 2009 by Pickett

[Guest]

Sum of all digits = 45 (1 to 9)

since each triplet is = 6e, sum of all digits = 25e + b+ d+ f +h; this means the max value of e = 3

E can not be 1 as it would make 7e = 7

For E = 2, if one of the other digits is 9, sum can not be made 14

So E must be 3

sum of each 2x2 matrix = 21

When one of the digits is 1, the sum of other two must be 21 - 1 -3 = 17;

so 1 must be paired wtih 9 and 8 and must be in one of the corners. Since 8 is one of the middle digits and 8+3 = 11, sum of other two must be 10; this can be done only for 4 &6 as tge other digits.

The rest is easy to figure out by eliminating

The digits are arranged as

1 8 4

9 3 6

2 7 5

I just saw that picket had answered both parts! Earlier I just saw part 1 at the top of his post and thought part 2 was yet to be done

Edited July 28, 2009 by DeeGee

[Guest]

The previous posts didn't post all of the solutions. In the first case, there are two solutions and in the second case there are at least two.

In the first case they are

345

927

618

and

587

321

496. It can be shown that these are in fact the only two (up to rotations and reflections).

In the second case they are

184

936

275

and

546

938

273.

(My guess is that these are the only two, but I didn't take the time to completely justify this hypothesis.)