Prime Number Equation

Let the equation $2a+3d=9999$ be equation $x$ . The equation $x$ 's RHS ( $9999$ ) is divisible by $3$ , as $9+9+9+9$ is divisible by $3$ , as $9$ is divisible by $3$ . Therefore since $d$ is an integer, $-3d$ must be divisible by $3$ . Subtracting $3d$ from both sides of equation $x$ , $2a+3d-3d=9999-3d$ . Eliminating the $3d$ and the $-3d$ from the LHS of this new equation, $2a=9999-3d$ . Let this new equation be equation $y$ . Since the equation $x$ 's RHS ( $9999$ ) is divisible by $3$ , since $-3d$ is divisible by 3, and since they are joined by the additive operator to form the RHS of equation $y$ , the RHS of equation $y$ must be divisible by 3, and so must the LHS of equation $y$ . Since $2a$ is divisible by $3$ , since $a$ is an integer, since $2$ is not divisible by $3$ , and since the $2$ and the $a$ are connected by a multiplicative operator, $a$ is divisible by $3$ . Since $a$ is a positive prime number and is divisible by $3$ , there is one distinct value of $a$ and only one distinct value of $a$ , as possible. There also must be a corresponding value of $d$ . There is one distinct value of $(a, b)$ and only one distinct value of $(a, b)$ . All the requirements are met, and this answer is checkable, as $a$ would equal $3$ , and therefore from substitution and simplifications from equation $x$ , $d=3331$ , and all the requirements are met. Therefore, the answer is $1$ .