This is called a math problem

Let the required number be $100x+10y+z$ , where $x, y, z$ are the digits of the number. Then $20y+2z=70+C$ . Since $z$ can't be greater than $9$ , therefore $y=3,z=\dfrac{C}{2}+5$ . Also, from the given conditions, $200x+1000z\geq {10000}$ (since the hundred's and thousand's place digits of the product are both zero). Since $z_{max}=9$ and $200x\leq {1800}$ , therefore $x=5$ and $z=9$ , and the required number is $5\times {100}+3\times {10}+9=\boxed {539}$