Problem (Number Theory):
Determine the number of quadruples of positive integers $(a, b, c, d)$ with $a<b<c<d$ that satisfy both of the following system of equations:

$$
\begin{aligned}
a c+a d+b c+b d & =2023 \\
a+b+c+d & =296
\end{aligned}
$$

Solution:
Factoring the first equation, we obtain

$$
a c+a d+b c+b d=a(c+d)+b(c+d)=(a+b)(c+d)
$$

We now have the equations

$$
\begin{aligned}
(a+b)(c+d) & =2023 \\
(a+b)+(c+d) & =296
\end{aligned}
$$

If we let $s=a+b$ and $t=c+d$, we obtain the equations

$$
\begin{aligned}
s t & =2023 \\
s+t & =296
\end{aligned}
$$

Noting that $s$ and $t$ are integers since $a, b, c$, and $d$ are integers, we look for divisor pairs of 2023 whose sum is 296 .

To find the divisors of 2023 , we first find its prime factorization:

$$
2023=7 \cdot 289=7 \cdot 17^{2}
$$

Therefore, the divisors of 2023 are 1, 7, 17, 119, 289, 2023.

This means that the divisor pairs of 2023 are

$$
2023=1 \cdot 2023=7 \cdot 289=17 \cdot 119
$$

The one divisor pair with a sum of 296 is 7 and 289. (Alternatively, we could have found these by substituting $t=206-s$ into $s t=2023$ and using the quadratic formula.)



Since $a<b<c<d$, then $a+b<c+d$ and so $s=a+b=7$ and $t=c+d=289$.

Since $a$ and $b$ are positive integers with $a<b$ and $a+b=7$, then the possible pairs $(a, b)$ are

$$
(a, b)=(1,6),(2,5),(3,4)
$$

We know that $c$ and $d$ are positive integers with $c<d$ and $c+d=289$, but also with $b<c<d$.

When $(a, b)=(1,6)$, this means that the possibilities are

$$
(c, d)=(7,282),(8,281),(9,280), \ldots,(143,146),(144,145)
$$

There are $144-7+1=138$ such pairs.

When $(a, b)=(2,5)$, the possibilities are

$$
(c, d)=(6,283),(7,282),(8,281),(9,280), \ldots,(143,146),(144,145)
$$

There are $138+1=139$ such pairs.

When $(a, b)=(3,4)$, the possibilities are

$$
(c, d)=(5,284),(6,283),(7,282),(8,281),(9,280), \ldots,(143,146),(144,145)
$$

There are $139+1=140$ such pairs.

In total, there are $138+139+140=417$ possible quadruples $(a, b, c, d)$.

Final answer: 417