968. Can 12 consecutive natural numbers be written in the cells of a rectangular table with 3 rows and 4 columns, arranging them so that:

a) the sum of the numbers in each row is divisible by 6;

b) the sum of the numbers in each column is divisible by 4;

c) the sum of the numbers in each column is divisible by 2, but not by 4?
$\triangle$ a) Let's take natural numbers from 1 to 12 and arrange them in a $3 \times 4$ table as follows.

Since the sum of the numbers in each row is divisible by 6, it is possible to arrange them in the table in the required manner.

| 1 | 2 | 3 | 6 |
| :---: | :---: | :---: | :---: |
| 4 | 5 | 7 | 8 |
| 9 | 10 | 11 | 12 |

(It can be proven that it is possible to arrange any 12 consecutive natural numbers in the cells of the table in the same way.)

b) To answer the next question of the problem, let's take any 12 consecutive natural numbers

$$
n, n+1, n+2, \ldots, n+11
$$

and find their sum:

$$
n+(n+1)+(n+2)+\ldots+(n+11)=6 \cdot(2 n+11)
$$

(check this!). If the sum of the numbers in each column were divisible by 4, then the sum of all the numbers in the table would also be divisible by 4. But this is not true. Therefore, it is impossible to arrange the numbers in the table in such a way.

c) Suppose the sum of the numbers in each column of the table is divisible by 2 but not by 4. Then this sum in the first, second, third, and fourth columns is respectively

$$
2 a, 2 b, 2 c, 2 d \text {, }
$$

where $a, b, c, d$ are odd numbers. Let's find the sum of all the numbers in the table.

$$
2 a+2 b+2 c+2 d=2(a+b+c+d)
$$

Since the sum $a+b+c+d$ is even as the sum of four odd numbers, it turns out again that the sum of all the numbers in the table is divisible by 4, which is impossible.

Answer: a) can; b) cannot; c) cannot.