$Ax+By+Cz=P$ , $0\le x,y,z$ and $200 \le \frac{C}{gcd(A,B,C)}$. For given $A,B,C,P$ find number of triplet $(x,y,z)$. Solution: \begin{align} 200 &\le \frac{C}{gcd(A,B,C)} \newline C &\ge 200\times gcd(A,B,C) \newline \therefore C &\ge 200 \end{align} Hence, we will rewrite the eqation as $Ax+By+Cz=P \Longrightarrow Ax+By=P-Cz=P'$ We will iterate over all possible values of $P'$(there will be maximum $\frac{10^{8}}{200}$), and for each value of $P'$ we will find number of pair $(x,y)$ that satisfy $Ax+By=P'$ , and sum them up.