Problem 720

Since $f(p)=17$, then $ap^{2}+bp+c=17$. Since $f(q)=17$, then $aq^{2}+bq+c=17$. Subtracting these two equations, we obtain $a(p^{2}-q^{2})+b(p-q)=0$. Since $p^{2}-q^{2}=(p-q)(p+q)$, this becomes $a(p-q)(p+q)+b(p-q)=0$. Since $p<q$, then $p-q \neq 0$, so we divide by $p-q$ to get $a(p+q)+b=0$. Since $f(p+q)=47$, then $a(p+q)^{2}+b(p+q)+c=47$ and so $(p+q)(a(p+q)+b)+c=47$. Since $a(p+q)+b=0$, then $(p+q)(0)+c=47$ which tells us that $c=47$. Since $ap^{2}+bp+c=17$, then $ap^{2}+bp=-30$ and so $p(ap+b)=-30$. Similarly, $q(aq+b)=-30$. Since $p$ and $q$ are prime numbers and $a$ and $b$ are integers, then $p$ and $q$ must be prime divisors of -30. We note that $30=2 \cdot 3 \cdot 5$ and also that $p$ and $q$ must be distinct. Since $p<q$, then $p=2$ and $q=3$, or $p=2$ and $q=5$, or $p=3$ and $q=5$. Alternatively, we could note that since $f(p)=f(q)=17$, then $f(p)-17=f(q)-17=0$. Therefore, $f(x)-17$ is a quadratic polynomial with roots $p$ and $q$, which means that we can write $f(x)-17=a(x-p)(x-q)$, since the quadratic polynomial has leading coefficient $a$. Since $f(p+q)=47$, then $f(p+q)-17=a(p+q-p)(p+q-q)$ which gives $47-17=aqp$ or $apq=30$. As above, $p=2$ and $q=3$, or $p=2$ and $q=5$, or $p=3$ and $q=5$. If $p=2$ and $q=3$, the equations $p(ap+b)=-30$ becomes $2(2a+b)=-30$ (or $2a+b=-15)$ and the equation $q(aq+b)=-30$ becomes $3(3a+b)=-30$ (or $3a+b=-10)$. Subtracting $2a+b=-15$ from $3a+b=-10$, we obtain $a=5$ (note that $a>0$) which gives $b=-15-2 \cdot 5=-25$. Therefore, $f(x)=5x^{2}-25x+47$. Since $pq=6$, then $f(pq)=5(6^{2})-25(6)+47=77$. If $p=2$ and $q=5$, we get $2a+b=-15$ and $5a+b=-6$. Subtracting the first of these from the second, we obtain $3a=9$ which gives $a=3$ (note that $a>0)$ and then $b=-15-2 \cdot 3=-21$. Therefore, $f(x)=3x^{2}-21x+47$. Since $pq=10$, then $f(pq)=3(10^{2})-21(10)+47=137$. If $p=3$ and $q=5$, we get $3a+b=-10$ and $5a+b=-6$. Subtracting the first of these from the second, we obtain $2a=4$ which gives $a=2$ (note that $a>0)$ and then $b=-10-3 \cdot 2=-16$. Therefore, $f(x)=2x^{2}-16x+47$. Since $pq=15$, then $f(pq)=2(15^{2})-16(15)+47=257$. The sum of these values of $f(pq)$ is $77+137+257=471$. The rightmost two digits of this integer are 71.