, 2025

We adopt the notation [row number, column number] to be equal to
the contents of a specific square in the grid.

Since $[1,1]$ is $1$ and adjacent numbers increase by a
fixed integer $a>0$ moving left to
right within each row, then $[1,2]$
is $1+a$, $[1,3]$ is $1+2a$, and so on to the end of the row
where $[1,8]$ is $1+7a$.

Similarly, adjacent numbers increase by a fixed integer $b>0$ moving top to bottom within each
column, and so the contents of each square in row 2 is $b$ greater than the adjacent square in row
1.

That is, $[2,1]$ is $1+b$, $[2,2]$ is $1+a+b$, $[2,3]$ is $1+2a+b$, and so on to the end of the row
where $[2,8]$ is $1+7a+b$.

Continuing in this way in row 3, we get $[3,1]$ is $1+2b$, $[3,2]$ is $1+a+2b$, $[3,3]$ is $1+2a+2b$, and so on to $[3,8]$ which is equal to $1+7a+2b$.

We continue this pattern in the table below, and include each of the
entries in column 5 since the focus of the question is on this
column.

1
$1+a$
$1+2a$
$1+3a$
$1+4a$
$1+5a$
$1+6a$
$1+7a$

$1+b$
$1+a+b$
$1+2a+b$
$1+3a+b$
$1+4a+b$
$1+5a+b$
$1+6a+b$
$1+7a+b$

$1+2b$
$1+a+2b$
$1+2a+2b$
$1+3a+2b$
$1+4a+2b$
$1+5a+2b$
$1+6a+2b$
$1+7a+2b$

$1+3b$
$1+a+3b$
$\vdots$
$\vdots$
$1+4a+3b$
$\vdots$
$\vdots$
$\vdots$

$1+4b$
$1+a+4b$

$1+4a+4b$

$1+5b$
$1+a+5b$

$1+4a+5b$

$1+6b$
$1+a+6b$

$1+4a+6b$

$1+7b$
$1+a+7b$

$1+4a+7b$

$1+7a+7b$

We are given that the number in the bottom right corner of the grid
is less than $75$, and so $1+7a+7b<75$ or $7a+7b<74$.

Dividing each term of this inequality by $7$, we get $\dfrac{7a}{7}+\dfrac{7b}{7}<\dfrac{74}{7}$
or $a+b<\dfrac{74}{7}$ and since
$a$ and $b$ are positive integers, then $a+b\leq10$.

Since $a+b\leq10$, then by
multiplying each term by $4$, we get
$4a+4b\leq40$.

In the table above, $[5,5]$ is $1+4a+4b$. Since $4a+4b\leq40$, then $1+4a+4b\leq41$ and so $[5,5]$ cannot equal $45$.

Further, since $b$ is a positive
integer, then each of the entries in column $5$ above row 5 is less than $1+4a+4b$ and thus cannot equal $45$.

Thus if $45$ appears in column 5 of
this grid, then it can only appear in rows 6, 7 and 8.

We begin by noting that $a$ and
$b$ are positive integers, and since
$a+b\leq10$, then $a\leq9$ and $b\leq9$.

If $[6,5]$ is 45, then $1+4a+5b=45$ or $4a+5b=44$.

If $a=1$, then $4\times1+5b=44$ or $5b=40$, and so $b=8$.

We confirm that when $a=1$ and $b=8$, then $a+b\leq10$ and so the number appearing in
the bottom right corner of the grid is less than $75$.

Thus, in the arithmetic grid with $a=1$ and $b=8$, $[6,5]$ is 45.

If $a=2$, then $4\times2+5b=44$ or $5b=36$ or $b=\dfrac{36}{5}$ (which is not an
integer), and so there is no such arithmetic grid in which $[6,5]$ is $45$ when $a=2$.

We could continue substituting the remaining values of $a$ from $3$ to $9$ to determine for which values of $a$, $4a+5b=44$ and $a+b\leq10$ and $b$ is a positive integer.

Alternately, we might notice that $4a$ is a multiple of $4$, as is $44$.

Thus, if $4a+5b=44$ then $5b$ must also be a multiple of $4$ and so $b$ must be a multiple of $4$.

The only remaining positive integer $b$ for which $b\leq9$ and $b$ is a multiple of 4 is $b=4$.

When $b=4$, we get $4a+5\times4=44$ or $4a=24$ and so $a=6$.

We again confirm that when $a=6$ and
$b=4$, then $a+b\leq10$ and so the number appearing in
the bottom right corner of the grid is less than $75$.

In the arithmetic grid with $a=6$ and
$b=4$, $[6,5]$ is $45$.

Thus, there are exactly $2$ such
grids in which $45$ appears in row 6,
column 5.

If $[7,5]$ is $45$, then $1+4a+6b=45$ or $4a+6b=44$ or $2a+3b=22$.

In this case, we similarly notice that $2a$ is a multiple of $2$, as is $22$.

Thus, if $2a+3b=22$ then $3b$ must also be a multiple of $2$ and so $b$ must be a multiple of $2$.

We proceed by checking values of $b$
which are equal to positive even integers.

When $b=2$, we get $2a+3\times2=22$ or $2a=16$ and so $a=8$.

In this case, $a+b\leq10$ and so the
number appearing in the bottom right corner is less than $75$.

In the arithmetic grid with $a=8$ and
$b=2$, $[7,5]$ is $45$.

When $b=4$, we get $2a+3\times4=22$ or $2a=10$ and so $a=5$.

In this case, $a+b\leq10$.

In the arithmetic grid with $a=5$ and
$b=4$, $[7,5]$ is $45$.

When $b=6$, we get $2a+3\times6=22$ or $2a=4$ and so $a=2$.

In this case, $a+b\leq10$.

In the arithmetic grid with $a=2$ and
$b=6$, $[7,5]$ is $45$.

When $b=8$, we get $2a+3\times8=22$ which is not possible
since $a>0$.

Thus, there are exactly $3$ such
grids in which $45$ appears in row 7,
column 5.

If $[8,5]$ is $45$, then $1+4a+7b=45$ or $4a+7b=44$.

We notice that $4a$ is a multiple of
$4$, as is $44$.

Thus, if $4a+7b=44$ then $7b$ must also be a multiple of $4$ and so $b$ must be a multiple of $4$.

We proceed by checking the two possible values of $b$, namely $b=4$ and $b=8$.

When $b=4$, we get $4a+7\times4=44$ or $4a=16$ and so $a=4$.

In this case, $a+b\leq10$ and so the
number appearing in the bottom right corner is less than $75$.

In the arithmetic grid with $a=4$ and
$b=4$, $[8,5]$ is $45$.

When $b=8$, we get $4a+7\times8=44$ which is not possible
since $a>0$.

Thus, there is exactly $1$ such grid
in which $45$ appears in row 8,
column 5, and so there are $2+3+1=6$
grids in total.