There are functions $f(x)$ with the following properties:

- $f(x)=a x^{2}+b x+c$ for some integers $a, b$ and $c$ with $a>0$, and
- $f(p)=f(q)=17$ and $f(p+q)=47$ for some prime numbers $p$ and $q$ with $p<q$. For each such function, the value of $f(p q)$ is calculated. The sum of all possible values of $f(p q)$ is $S$. What are the rightmost two digits of $S$ ?
Since $f(p)=17$, then $a p^{2}+b p+c=17$.

Since $f(q)=17$, then $a q^{2}+b q+c=17$.

Subtracting these two equations, we obtain $a\left(p^{2}-q^{2}\right)+b(p-q)=0$.

Since $p^{2}-q^{2}=(p-q)(p+q)$, this becomes $a(p-q)(p+q)+b(p-q)=0$.

Since $p0$ ) which gives $b=-15-2 \cdot 5=-25$.

Therefore, $f(x)=5 x^{2}-25 x+47$.

Since $p q=6$, then $f(p q)=5\left(6^{2}\right)-25(6)+47=77$.

If $p=2$ and $q=5$, we get $2 a+b=-15$ and $5 a+b=-6$.

Subtracting the first of these from the second, we obtain $3 a=9$ which gives $a=3$ (note that $a>0)$ and then $b=-15-2 \cdot 3=-21$.

Therefore, $f(x)=3 x^{2}-21 x+47$.

Since $p q=10$, then $f(p q)=3\left(10^{2}\right)-21(10)+47=137$.

If $p=3$ and $q=5$, we get $3 a+b=-10$ and $5 a+b=-6$.

Subtracting the first of these from the second, we obtain $2 a=4$ which gives $a=2$ (note that $a>0)$ and then $b=-10-3 \cdot 2=-16$.

Therefore, $f(x)=2 x^{2}-16 x+47$.

Since $p q=15$, then $f(p q)=2\left(15^{2}\right)-16(15)+47=257$.

The sum of these values of $f(p q)$ is $77+137+257=471$.

The rightmost two digits of this integer are 71.

ANSWER: 71