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:

  1. 29
  2. 15
  3. 17
  4. 28
  5. 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:


Which appears as a choice, but my book says this is not the answer. What could I be doing wrong? Did I overlook something?

  • 4
    $\begingroup$ I think your answer is correct, as well as your solving path. There could be typos in the book itself. What's the answer provided by the book? $\endgroup$
    – Bubbler
    Nov 6, 2020 at 3:08
  • $\begingroup$ @Bubbler I'm sorry for the late reply buddy. But I'm starting to believe that there might be a typo in the answers sheet of the book. The book mentions the answer to be $14$ I have no idea how did they got to there. Maybe you want to add a clarification or an answer reassuring that my strategy was right?. $\endgroup$ Nov 8, 2020 at 2:35

1 Answer 1


As I already commented, I think your answer is indeed correct (and the book might have some typos there, likely in the answer sheet).

The 5 can't be at the center because then it would border with all of 6, 7, 8, and 9, which sum to 30.

$$ \begin{array}{|c|c|c|} \hline 1&?&3 \\ \hline ?&5&?\\ \hline 2&?&4\\ \hline \end{array} $$

So the 5 must be on one of four sides, sharing edges with two numbers $1\le a,b\le4$ (two adjacent corners) and one number $c\ge 6$ (the center).

We're given that $a+b+c=9$, which can only be satisfied when we choose the minimum possible for all three ($\{a,b\}=\{1,2\}, c=6$). So the 5 must be on the left edge, bordering with a 6.

$$ \begin{array}{|c|c|c|} \hline 1&?&3 \\ \hline 5&6&?\\ \hline 2&?&4\\ \hline \end{array} $$

For the remaining numbers (7, 8, 9), it doesn't matter how we place them as all of them will be adjacent to 6 anyway. Therefore the requested sum is $5+7+8+9 = 29$.


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