The integers 1 to 6 are to be inserted into the grid shown. No two integers that differ by 1 may be in squares that share an edge. If the 1 is inserted as shown, how many different integers can be placed in the box labelled $x$ ?
(A) 1
(B) 3
(C) 5
(D) 0
(E) 2

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-040.jpg?height=339&width=179&top_left_y=2107&top_left_x=1342)
We label the remaining unknown entries as $a, b, c, d$, as shown.

Now $a, b, c, d, x$ must equal $2,3,4,5,6$ in some order, with the restriction that no two integers that differ by 1 may be in squares that share an edge.

In particular, we see first that $a \neq 2$. Therefore, $a$ can equal $3,4,5$, or 6 .

If $a=3$, then neither $b$ nor $d$ can equal 2 or 4 .

Thus, $b$ and $d$ equal 5 and 6 in some order.

In this case, $c$ cannot be 4 (since one of $b$ and $d$ is 5) so $c=2$ and so $x=4$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-082.jpg?height=350&width=182&top_left_y=183&top_left_x=1687)

If $a=4$, then neither $b$ nor $d$ can equal 3 or 5 .

Thus, $b$ and $d$ equal 2 and 6 in some order.

In this case, $c$ cannot equal 3 or 5 , since it is adjacent to 2 and 6 , so this case is not possible.

If $a=5$, then neither $b$ nor $d$ can equal 4 or 6 .

Thus, $b$ and $d$ equal 2 and 3 in some order.

In this case, $c$ cannot be 4 (since one of $b$ and $d$ is 3 ) so $c=6$ and so $x=4$.

If $a=6$, then neither $b$ nor $d$ can equal 5 .

Thus, $b$ and $d$ equal two of 2,3 or 4 in some order.

If $b$ and $d$ are 2 and 4 , then $c$ cannot be 3 or 5 (the remaining numbers), so there is no solution.

If $b$ and $d$ are 3 and 4 , then $c$ cannot be 2 or 5 (the remaining numbers), so there is no solution. If $b$ and $d$ are 2 and 3 , then we can have $c=5$, in which case we must have $x=4$, which is not possible.

Having considered all possible cases, we see that there is only one possible value for $x$, namely $x=4$.

ANSWER: (A)