Take Twelve to Total Twenty Two

Question

Without repeating the numbers 1 to 12,

Fill the grid below to total 22 in both directions.

$$\def\X{\smash{\rlap{\Space{7pt}{0px}{0px}\llap{\Huge{\times}}}}} \Large\begin{array}{|c|c|c|c|}\hline &&5&\\\hline \rlap{10}~~~&\X&\X&\\[-2px]\hline &\X&\X&7~\\[-5px]\hline &9&&\\[-10pt]\hline \end{array}$$

Answer 1

Answer

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

Explanation

In the column containing 10 in order to make 22, the remaining three should be such that the sum is even. So we can have 2 odd and 1 even or 3 even numbers.
In the rows with 5 and 9 and in the column with 7, in order to make the sum even, we need 3 odd or 1 odd and 2 even numbers. But we have only 3 odd numbers left.
So I decided to place all the three in the right most column. So the left most column should have all even and the sum not exceeding 22.
This made me to place 2,4,6 in the left most column.
At last the numbers were to be arranged such that the sum is 22

Answer 2

TL;DR

There are two possible solutions.

Preliminary deductions

Left column:

22-10=12, so we need three numbers summing to 12, without using 5,7,9,10. Those three numbers must be $1,3,8$ or $2,4,6$.

Bottom row:

22-9=13, so we need three numbers summing to 13, without using 5,7,9,10. Those three numbers must be $1,4,8$ or $2,3,8$ or $3,4,6$.

In particular, the numbers

11 and 12 cannot appear in the left column or the bottom row. Clearly we can't have both 11 and 12 in one row/column, so one of them must be just left of 5 and the other one just above 7.

Option 1

Let's assume

11 is next to 5 and 12 is above 7.

Top row:

22-11-5=6, so the last two numbers must be $2,4$.

Right column:

22-12-7=3, so the last two numbers must be $1,2$.

So we have

4 11 5 2
10 X X 12
. X X 7
. 9 . 1

The remaining numbers are

$3,6,8$, but no two of these sum to 12 (for the bottom row) or 8 (for the left column), so there is no solution here.

Option 2

Let's assume

12 is next to 5 and 11 is above 7.

Top row:

22-12-5=5, so the last two numbers must be $2,3$ or $1,4$.

Right column:

22-11-7=4, so the last two numbers must be $1,3$.

So we have one of the following two possibilities:

4 12 5 1
10 X X 11
. X X 7
. 9 . 3


2 12 5 3
10 X X 11
. X X 7
. 9 . 1

The remaining numbers are respectively $2,6,8$ or $4,6,8$, so the complete grid is one of the following two:

4 12 5 1
10 X X 11
6 X X 7
2 9 8 3


2 12 5 3
10 X X 11
6 X X 7
4 9 8 1