CEMC Cayley, 2022

Suppose that Andreas, Boyu, Callista, and Diane choose the
numbers $a$, $b$, $c$, and $d$, respectively.

There are 9 choices for each of $a$,
$b$, $c$, and $d$, so the total number of quadruples
$(a,b,c,d)$ of choices is $9^4 = 6561$.

Among the 9 choices, 5 are odd ($1,3,5,7,9$) and 4 are even ($2,4,6,8$).

If there are $N$ quadruples $(a,b,c,d)$ with $a+b+c+d$ even (that is, with the sum of
their choices even), then the probability that the sum of their four
integers is even is $\dfrac{N}{6561}$, which is in the desired
form.

Therefore, we count the number of quadruples $(a,b,c,d)$ with $a+b+c+d$ even.

Among the four integer $a$, $b$, $c$, $d$, either 0, 1, 2, 3, or 4 of these
integers are even, with the remaining integers odd.

If 0 of $a$, $b$, $c$, $d$
are even and 4 are odd, their sum is even.

If 1 of $a$, $b$, $c$, $d$
is even and 3 are odd, their sum is odd.

If 2 of $a$, $b$, $c$, $d$
are even and 2 are odd, their sum is even.

If 3 of $a$, $b$, $c$, $d$
are even and 1 is odd, their sum is odd.

If 4 of $a$, $b$, $c$, $d$
are even and 0 are odd, their sum is even.

Therefore, we need to count the number of quadruples $(a,b,c,d)$ with 0, 2 or 4 even parts.

If 0 are even and 4 are odd, there are 5 choices for each of the
parts, and so there are $5^4 = 625$
such quadruples.

If 4 are even and 0 are odd, there are 4 choices for each of the parts,
and so there are $4^4 = 256$ such
quadruples.

If 2 are even and 2 are odd, there are 4 choices for each of the even
parts and 5 choices for each of the odd parts, and 6 pairs of locations
for the even integers ($ab, ac, ad, bc, bd, cd$) with the odd integers put in the remaining two locations
after the locations of the even integers are chosen. Thus, there are
$4^2 \cdot 5^2 \cdot 6 = 2400$ such
quadruples.

In total, this means that there are $625 + 256 + 2400 = 3281$ quadruples and so $N = 3281$.

The sum of the squares of the digits of $N$ is equal to $3^2 + 2^2 + 8^2 + 1^2 = 9 + 4 + 64 + 1 = 78$.