Let $\text{AP}(a;d)$ denote the set of all the terms of an infinite arithmetic progression with first term $a$ and common difference $d>0$. If $\text{AP}(1;3)\cap \text{AP}(2;5)\cap \text{AP}(3;7)=\text{AP}(a;d)$ then $a+d$ equals _____.

$\text{Given: } \text{AP}(a;d)=\{a,\ a+d,\ a+2d,\ a+3d,\ \dots\},\quad d>0$ $\text{AP}(1;3),\ \text{AP}(2;5),\ \text{AP}(3;7)$ $\text{AP}(1;3)\cap \text{AP}(2;5)\cap \text{AP}(3;7)=\text{AP}(a;d)$ $\text{To find: } a+d$ $\text{AP}(1;3):\ n \equiv 1 \pmod 3$ $\text{AP}(2;5):\ n \equiv 2 \pmod 5$ $\text{AP}(3;7):\ n \equiv 3 \pmod 7$ $\text{AP}(1;3)=\{1,\ 4,\ 7,\ 10,\ 13,\ \dots\}$ $\text{AP}(2;5)=\{2,\ 7,\ 12,\ 17,\ \dots\}$ $\text{AP}(3;7)=\{3,\ 10,\ 17,\ 24,\ \dots\}$ $52 = 1 + 3(17) = 2 + 5(10) = 3 + 7(7)$ $\text{Plan: }1.\ \text{each AP} \to \text{one congruence}$ $2.\ \text{solve pairwise, then mod } 3\cdot 5\cdot 7 = 105$ $3.\ \text{read } a,\ d \Rightarrow a+d$ $n \equiv 1 \pmod 3 \Rightarrow n = 1 + 3k$ $1 + 3k \equiv 2 \pmod 5$ $3k \equiv 1 \pmod 5$ $3k \equiv 1 \pmod 5 \Rightarrow k \equiv 2 \pmod 5$ $k = 2 + 5m$ $n = 1 + 3(2 + 5m) = 7 + 15m$ $n \equiv 7 \pmod{15}$ $7 + 15m \equiv 3 \pmod 7$ $15m \equiv -4 \equiv 3 \pmod 7$ $15 \equiv 1 \pmod 7 \Rightarrow m \equiv 3 \pmod 7$ $m = 3 + 7t$ $n = 7 + 15(3 + 7t) = 52 + 105t$ $n \equiv 52 \pmod{105}$ $\text{AP}(a;d) = \text{AP}(52;105) \Rightarrow a = 52,\quad d = 105$ $a + d = 52 + 105$ $a + d = 157$