The problem is as follows:
I am placing the numbers from 1 to 9 in the squares of a 3 x 3 board. I start by placing the numbers 1, 2, 3 and 4 as shown in the figure. For the number 5, the sum of the numbers in the adjacent boxes (which have a common side with the 5) is equal to 9. What is the sum of the numbers adjacent to 6?
$$\begin{array}{|c|c|c|} \hline 1&\,\,&3 \\ \hline \,\,&\,\,&\,\,\\ \hline 2&\,\,&4\\ \hline \end{array}$$
The choices given in my book are as follows:
- 29
- 15
- 17
- 28
- 14
I found this problem in my puzzles book Reason and Logic from the 2000s. The question seems to be adapted from a reprinted copy of Martin Gardner's 50's book on Recreational puzzles.
The nature of this problem is told as a narration from the author. The way how I understood what it was intended was as follows:
The only available numbers to use are: 5, 6, 7, 8 and 9.
But since it mentions that the empty spaces which are contiguous to 5 will add up to 9, will leave the only choice (as I could spot on) to be this one:
$$ \begin{array}{|c|c|c|} \hline 1&\,\,&3 \\ \hline 5&6&\,\,\\ \hline 2&\,\,&4\\ \hline \end{array}$$
As $1+6+2=9$ Hence this leaves up the other number unused available, with those being: 7, 8 and 9.
Since it doesn't say anything else about them, I ordered them this way:
\begin{array}{|c|c|c|} \hline 1&7&3 \\ \hline 5&6&8\\ \hline 2&9&4\\ \hline \end{array}
As the problem requests to add up the digits contiguous to 6 these will:
$7+8+9+5=29$
Which appears as a choice, but my book says this is not the answer. What could I be doing wrong? Did I overlook something?