Michael's divisible grids

Name the squares in the top row from left to right A, B and C.

Name the squares in the bottom row from left to right D, E and F.

We can choose freely what to put in squares A and B; 12 options each.

Now A+B is either even or odd, so square C must be even or odd; 6 options.

From now on, consider all squares modulo 3.

Now A is 0, 1 or 2, so D has to be 0, 2 or 1. Thus there are 4 options for square D

Square B is 0, 1 or 2, so E has to be 0, 2 or 1. Thus there are 4 options for square E.

Square C is 0, 1 or 2, so F has to be 0, 2 or 1. This gives 4 options for square F, but F also needs to be even or odd (depending on D+E being even or odd). This leaves 2 options for square F.

In total we have $N = 12*12*6*4*4*2 = 27648$

Let each element in first row have 12 possibilities. Corresponding element in 2nd row will have 4 possibilities to satisfy divisibility by 3 condition. These include cases with sum in each row being OO/OE/EO/EE which are equiprobable.(O: Odd, E: even) So answer (12 12 12 4 4*4)/4 = 27648

Suppose the numbers in the grid are $\begin{array}{|c|c|c|} \hline a & b & c \\ \hline d & e & f \\ \hline \end{array}$ Then $a$ and $b$ can be chosen freely, so there are $12$ choices for each of these. To get the top row sum right, $c$ is specified modulo $2$ , and so there are $6$ choices for $c$ . To get the first two columns right, $d$ and $e$ are specified modulo $3$ , and hence there are $4$ choices for each of these.

To get the bottom row sum right, $f$ is specified modulo $2$ . To get the right-hand column sum right, $f$ is also specified modulo $3$ . By the Chinese Remainder Theorem, $f$ is uniquely determined modulo $6$ , and hence there are $2$ choices for $f$ . Thus there are $12 \times 12 \times 6 \times 4 \times 4 \times 2 \; = \; 27648$ possible grids, and so the answer is $648$ .

Let us first simplify this problem until it is more manageable. Based on the conditions required for an allowed rectangle, the only relevant information about an integer is its value mod 6. Thus we will reduce our integers to the range from 0 to 5.

Now we will look for a way to define a 2x3 rectangle without filling it out completely. Notice that if we almost fill a row but leave one square empty, and then pick an integer above (or below) that empty square, then the empty square can only take on one value.

In the following diagram, let X denote an entirely independent square, O an entirely dependent square, and V a square that can only take on two values.

Thus there are $2 \cdot 6^3$ possibilities under modulo 6. However, each integer in the rectangle can be transformed into one other integer in the range of 0 to 11 without voiding its validity. This gives us our final answer of,

$2 \cdot 6^3 \cdot 2^6 = 27648 \equiv \boxed{648} \pmod{1000}$

For this question, I shall choose to differentiate the possible numbers by parity, that is, whether they are even or odd.

For each row, there are either $2$ odd numbers and one even number, or all $3$ being even numbers since their sum is to be divisible by $2$ . Thus, with the order of the terms differentiated by parity not mattering (odd, even, odd is considered distinct compared to even, odd, odd), the total possible combinations for parity of the two rows in terms of parity will be $4*4$ (like even, even, even for the top row and odd, even, odd for the bottom row).

Furthermore, once we define the terms in the rows by parity for say, the top row (assuming w.l.o.g.), we will have $6$ possibilities for each of the cells in the row (i.e. for the case odd,even, odd: $6$ possibilities for the odd number, $6$ possibilities for the even number, $6$ possibilities for the second odd number, etc.). Consequently, with the bottom row's cell directly below the corresponding cell in the top row for each of the $4$ possible combinations in can undergo, there will be thus exactly $2$ numbers that can be placed in that particular cell for the row's $3$ cells, since the specified parity of the cell limits it to $6$ possibilities, while their being taken $(\text{mod 3})$ would restrict it further to only $\frac{1}{3}$ of the said possibilities, thus giving $2$ possibilities. Three cells in the top row and their corresponding three cells in the bottom row as differentiated by parity thus give us $6^3*2^3$ possibilities (For each of the $4*4$ cases above, of course.)

As an explicit example: say the odd number in the one of the cells of the top row is $11$ (out of $6$ possibilities), and below that cell it has to be an even number. For their sum to be divisible by $3$ , we have only $2$ possibilities for the value in the corresponding cell in the bottom row (either $1$ or $7$ ). The cases for the other two cells in the top row (and their corresponding cells in the bottom row) follow analogously.

Altogether we have $4*4*6^3*2^3=27648$ possible ways to fill the grid, with the last 3 digits being $\fbox{648}$