18. Let $f(x)$ be the polynomial $\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)\left(x-a_{4}\right)\left(x-a_{5}\right)$ where $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ are distinct integers. Given that $f(104)=2012$, evaluate $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}$.
18. Answer: 17
The prime factorization of 2012 is $2^{2}$ - 503. Let $b=104$. If $a_{i}$ are distinct, so are $b-a_{i}$, i.e. $\left(b-a_{1}\right),\left(b-a_{2}\right),\left(b-a_{3}\right),\left(b-a_{4}\right)$ and $\left(b-a_{5}\right)$ must be exactly the integers $1-12-2503$. Summing up, we have
$$
5(104)-\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)=1-1+2-2+503
$$
ie. $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=17$.