CEMC Fermat, 2024

We note that $$\begin{aligned} ac + bd - ad - bc &= ac - ad - bc + bd\\ & = a(c-d) - b(c-d) \\ & = (a-b)(c-d)\end{aligned}$$ Since each of $a$, $b$, $c$, $d$
is taken from the set $\{1,2,3,4,5,6,7,8,9,10\}$, then $a - b \leq 9$ since the greatest possible
difference between two numbers in the set is $9$.

Similarly, $c-d \leq 9$.

Now, if $a-b = 9$, we must have $a = 10$ and $b = 1$.

In this case, $c$ and $d$ come from the set $\{2, 3, 4, 5, 6, 7, 8, 9\}$ and so $c-d \leq 7$.

Therefore, if $a-b = 9$, we have
$(a-b)(c-d) \leq 9 \cdot 7 = 63$.

If $a - b = 8$, then either $a = 9$ and $b = 1$, or $a = 10$ and $b = 2$.

In both cases, we cannot have $c - d = 9$ but we could have $c - d = 8$ by taking the other of these two pairs with a difference of
$8$.

Thus, if $a - b = 8$, we have $(a-b)(c-d) \leq 8 \cdot 8 = 64$.

Finally, if $a - b \leq 7$, the
original restriction $c - d \leq 9$
tells us that $(a-b)(c-d) \leq 7 \cdot 9 = 63$.

In summary, the greatest possible value for $ac + bd - ad - bc$ is 64 which occurs, for
example, when $a = 9$, $b = 1$, $c = 10$, and $d = 2$.