[Algebra]
Problem: Consider the system of equations:

$$
\begin{aligned}
c+d & =2000 \\
\frac{c}{d} & =k
\end{aligned}
$$

Determine the number of integers $k$ with $k \geq 0$ for which there is at least one pair of integers $(c, d)$ that is a solution to the system.

Solution: From the second equation, we note that $d \neq 0$.

Rearranging this second equation, we obtain $c=k d$.

Substituting into the first equation, we obtain $k d+d=2000$ or $(k+1) d=2000$.

Since $k \geq 0$, note that $k+1 \geq 1$.

This means that if $(c, d)$ is a solution, then $k+1$ is a divisor of 2000 .

Also, if $k+1$ is a divisor of 2000 , then the equation $(k+1) d=2000$ gives us an integer value of $d$ (which is non-zero) from which we can find an integer value of $c$ using the first equation.

Therefore, the values of $k$ that we want to count correspond to the positive divisors of 2000.

Since $2000=10 \cdot 10 \cdot 20=2^{4} \cdot 5^{3}$, then 2000 has $(4+1)(3+1)=20$ positive divisors.

This comes from the fact that if $p$ and $q$ are distinct prime numbers then the positive integer $p^{a} \cdot q^{b}$ has $(a+1)(b+1)$ positive divisors.

We could list these divisors as

$$
1,2,4,5,8,10,16,20,25,40,50,80,100,125,200,250,400,500,1000,2000
$$



if we did not know the earlier formula.

Since 2000 has 20 positive divisors, then there are 20 values of $k$ for which the system of equations has at least one integer solution.

For example, if $k+1=8$, then $k=7$. This gives the system $c+d=2000$ and $\frac{c}{d}=7$ which has solution $(c, d)=(1750,250)$.

Final answer: 20