Posted by **Mark** on Monday, July 8, 2013 at 6:58pm.

The squares of a 3×3 grid are filled with non-negative integers such that the sum of each row and the sum of each column is 7. How many different ways can the squares be filled? The numbers in each grid square does not need to be distinct. Rotations and reflections are distinct arrangements.

- Math (please help me Steve) -
**MathMate**, Tuesday, July 9, 2013 at 6:51am
Here's a lower bound solution. You may want to investigate other possibilities.

The number 7 can be partitioned into 3 ascending non-negative integers in 8 ways, namely:

007#

016

025

034

115#

124

133#

233#

There are 3 permutations of partitions (indicated #) with non-distinct digits and 6 permutations of those with distinct digits for a total of

4*3+4*6=36 permutations of non-negative integers whose sum is 7.

Each of these permutations can make a 3x3 grid as follows by putting a permutation as the first line, then rotate the digits to the left:

124

241

412

We can also make another grid by rotating them to the right, as follows:

124

412

241

This doubles the number of grids for a total of

36*2=72 grids.

- Math (please help me Steve) -
**math**, Wednesday, July 10, 2013 at 6:36am
what about this case, i didnt enumerated it

1 2 4

0 5 2

6 0 1

- Math (please help me Steve) -
**exactly**, Sunday, September 22, 2013 at 1:47am
Hint: you need these:

(1) a,b,c,d are non-negative integers

(2) 7−a−b≥0⇔a+b≤7

(3) 7−c−d≥0⇔c+d≤7

(4) 7−a−c≥0⇔a+c≤7

(5) 7−b−d≥0⇔b+d≤7

(6) a+b+c+d−7≥0⇔a+b+c+d≥7

Try to count how many satisfies these conditions (:

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