6
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Without repeating the numbers 1 to 12,

Fill the grid below to total 22 in both directions.

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

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  • $\begingroup$ I replaced your photographed hand-drawn grid with a typed rendition of the same grid. If you don't like the new version, feel free to roll back my edit. $\endgroup$ – Rand al'Thor Jun 15 at 11:44
  • $\begingroup$ Thx..looks much better..I am not well versed in math Jax and other computer generated ones..also, I just sketch it in my puzzle book and is easier to take a picture..especially some complex ones $\endgroup$ – Uvc Jun 15 at 11:48
  • $\begingroup$ No problem! :-) $\endgroup$ – Rand al'Thor Jun 15 at 11:48
  • $\begingroup$ I replaced your typed rendition of the same grid with a MathJax version that looks much closer to the hand-drawn original. If you don't like the new new version, feel free to roll back my edit. $\endgroup$ – Rubio Jun 15 at 21:48
  • $\begingroup$ This looks nice and compact..thx $\endgroup$ – Uvc Jun 15 at 22:24
4
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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

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  • 1
    $\begingroup$ That's only one of three possibilities, I think. Also, not including explanation let you ninja me by under a minute :-D $\endgroup$ – Rand al'Thor Jun 15 at 9:38
  • $\begingroup$ 😭😭😭😭😭😭😭😭 $\endgroup$ – Ak19 Jun 15 at 9:48
  • $\begingroup$ Don't worry, +1. It is quicker to make some assumptions than to examine all possibilities exhaustively :-) $\endgroup$ – Rand al'Thor Jun 15 at 9:54
  • $\begingroup$ 👍👍👍😄😄😄😄😄😄 $\endgroup$ – Ak19 Jun 15 at 9:56
5
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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

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  • $\begingroup$ Your option 1 solution doesn't add up to 22 on the top or right. $\endgroup$ – JS1 Jun 15 at 11:11
  • $\begingroup$ @JS1 Oops! I got my 1 and 4 muddled up. Fixed, thanks. $\endgroup$ – Rand al'Thor Jun 15 at 11:15

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