Ramanujan-Diophantus

Consider $\pmod{17}$ ,then $0+12D\equiv 12 \pmod{17}\\ \Longrightarrow D\equiv 1\pmod{17}\\ \Longrightarrow D=17X+1\ \ \ \ \ [X \in \mathbb{Z}^{+} + {0}]\\ \Longrightarrow 17R+17*29X=1700\\ \Longrightarrow R+29X=100$ ,from here one can easily find the solutions as $29*4>100$ ,so one has to check only from $0-3$ .

Though Adarsh has posted the elegant method , this one is for the ones who have just started deciphering Diophantine equations :).

Since $gcd(17,29) = 1$ and $1 | 1729$ , there are integral solutions possible for this equation.

We use the Euclidean Algorithm :

$1 = 5-2(2) \\\Rightarrow 1=5-2(12-5(2)) \\\Rightarrow 1= 5(5)-2(12) \\ \Rightarrow 1=5(17-12) - 2(12) \\\Rightarrow 1= 5(17) - 7(12) \\\Rightarrow 1=5(17) - 7(29-17) \\\Rightarrow 1=12(17) - 7(29) \\ \Rightarrow 1729 = 20748(17) - 12103(29)$

So we have the following solutions :

$R=20748-29\delta \ , \ D=-12103+17\delta \ , \ \forall \delta \in \mathbb{Z}$

Since we want positive integer solutions ,

$20748-29\delta > 0 \ , \ -12103+17\delta>0 \\ \delta<\dfrac{20748}{29} \approx 715.4 \ , \ \delta>\dfrac{12103}{17} \approx 711.9 \\ \Rightarrow \delta \in \{712,713,714,715\}$

When $\delta=712$ we have a pair $(100,1)$ .

When $\delta=713$ we have a pair $(71,18)$ .

When $\delta=714$ we have a pair $(42,35)$ .

When $\delta=715$ we have a pair $(13,52)$ .

Hence required sum $=100+1+71+18+42+35+13+52 = \Large\boxed{332}$

Credits to Pi Han Goh for spotting my mistake.

We can immediately see we have the solution $R=100, D=1$ .

Now from properties of Diophantine Equations we know all solutions are of the form $(100-29k, 1+17k)$ since $17$ and $29$ are relatively prime. Since we need $R$ and $D$ to be positive, this holds for $k = 0$ to $k = 3$ , and for each we have the sum $101-12k$ .

Thus we have $4*101 - (0+1+2+3)12 = 404 - 72 = 332$ .