[tex]v[/tex] is an integer such that
[tex]\begin{cases}v\equiv2\equiv0\pmod2\\v\equiv2\pmod3\\v\equiv2\pmod5\end{cases}[/tex]
Since 2, 3, and 5 are mutually coprime, you can use the Chinese remainder theorem directly.
Suppose
[tex]v=2\cdot3\cdot5+2\cdot2\cdot5+2\cdot2\cdot3[/tex]
Then taken mod 2, the last two products vanish and you're left with a remainder of 0 (since 2 divides 2*3*5);
taken mod 3, the first and last products vanish and you're left with a remainder of 2 (since 2*2*5 = 20, and 20 = 2 mod 3);
and taken mod 5, the first two products vanish and you're left with a remainder of 2 (since 2*2*3 = 12, and 12 = 2 mod 5).
So we have [tex]v=62[/tex], and [tex]62\pmod{2\cdot3\cdot5}\equiv62\pmod{30}\equiv2\pmod{30}[/tex], so that any integer of the form [tex]v=30n+2[/tex] satisfies the system of congruences, the largest of which is 242.
