, 2020

Solution:
We first solve for $a$. Note that $3003=3 \cdot 7 \cdot 11 \cdot 13$. We have $3003=\binom{a}{2}=\frac{a(a-1)}{2} \approx \frac{a^{2}}{2}$. This means we can estimate $a \approx \sqrt{3003 \cdot 2}$, so $a$ is a little less than $80$. Furthermore, $11 \mid 2 \cdot 3003=a(a-1)$, meaning one of $a$ or $a-1$ must be divisible by $11$. Thus, either $a=77$ or $a=78$. Conveniently, $13 \mid 78$, so we get $a=78$ and we can verify that $\binom{78}{2}=3003$.

We solve for $b<15-b$ satisfying $\binom{15}{b}=3003$. Because $\binom{15}{b}=\frac{15!}{b!(15-b)!}$ is divisible by $11$, we must have $b \geq 5$. But we're given $\binom{14}{6}=3003$, so $b<6$. We conclude that $b=5$, and it follows that $a+b(15-b)=128$.