Chinese Remainder Theorem

Let $A$ is the smallest positive integer that fulfill the conditions. $A \equiv 1 \pmod 3 \Rightarrow 2A+1 \equiv 3 \equiv 0 \pmod 3$ . By doing the same thing we have $2A+1 \equiv 0 \pmod 5$ and $2A+1 \equiv 0 \pmod 7$ . To find the smallest value of $A$ , let $2A+1 = lcm(3,5,7) = 105 \Rightarrow A = \boxed{52}$ .