Logic with numbers Nick wrote each of the numbers from 1 to 9 in the cells of the 3x3 table below. Only 4 of the numbers can be seen in the figure. Nick noticed that for the number 5, the sum of the numbers in the neighboring cells is equal to 13 (neighboring cells are cells that share a side). He noticed that the same is also true for the number 6. Which number did Nick write in the shaded cell?

 A: We have the following figure 
We know that for the numbers $5$ and $6$, the sum of the numbers in the neighboring cells is equal to 13. 


*

*When $c_1=5$ or $6$ then it would have to hold that $1+2+c_3=13 \Rightarrow c_3=10$. This cannot be true since the greatest number is $9$. 

*When $c_5=5$ or $6$ then it would have to hold that $4+3+c_3=13 \Rightarrow c_3=6$. 
So, it must be $c_5=5$. That means that it must hold $4+6+3=13$. This is true. 
The sum of the neighboring cells of $c_3=6$ is $c_1+c_2+c_3+c_4=5+7+8+9>13$.   
($7,8,9$ are the left numbers that have to be in these cells.) 
This case is not possible. 

*When $c_4=5$ or $6$ then it would have to hold that $2+3+c_3=13 \Rightarrow c_3=8$. 
Without loss of generality, we suppose that $c_4=5$. 
Now we have to look in which cell is the number $6$. 
We check the cell $c_5$. When $c_5=6$, then it must hold that $4+8+3=13$. Which is not true. 
So, the only possibility is that $c_2=6$. Then it must hold that $1+8+4=13$, which is true. 
Therefore, at the cells $c_2,c_4$ there are the numbers $5,6$ and at the shaded cell $c_3$ there is the number $8$. 
A: The numbers $5$ and $6$ cannot be in the upper middle square, since then the shaded square has to contain $13-1-2=10$, and that's not possible. If the number $6$ is in the bottom middle square, then the shaded square contains $13-3-4=6$, which is not possible. If the number 5 is in the bottom middle square, then the shaded square contains $6$, but the neighbours then are $5+7+8+9=29$ (the only ones left) and so this is also impossible. If 5 is in the shaded square, then its neighbours add up to $6+7+8+9=30$, so also impossible. Thus, $5$ and $6$ must be on the sides, so the shaded square must contain $13-1-4=13-2-3=8$.
Hope this helped!
A: The number in the middle, shaded square abuts four numbers whose sum is at least $5+6+7+8=26$, so cannot be either the $5$ or the $6$.  Hence the $5$ and $6$ each go in a side square, abutting whatever's in the middle and two corner squares.  Since the total abutting sums both equal $13$, the corner abutting sums for the $5$ and $6$ must be equal.  This only happens for the corner sums $1+4$ and $2+3$ (implying the $5$ and $6$ occupy the sides squares on the left and right, in some order), which means the middle, shaded square contains the $8$.
