Question

If v is a three digit number less than 250 that has a remainder of 2 when divided by 2, 3 or 5, what is a possible value for v?

Answer 1

`v` is an integer such that

`\begin{cases}v\equiv2\equiv0\pmod2\\v\equiv2\pmod3\\v\equiv2\pmod5\end{cases}`

Since 2, 3, and 5 are mutually coprime, you can use the Chinese remainder theorem directly.

Suppose

`v=2\cdot3\cdot5+2\cdot2\cdot5+2\cdot2\cdot3`

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
`v=62`, and
`62\pmod{2\cdot3\cdot5}\equiv62\pmod{30}\equiv2\pmod{30}`, so that any integer of the form
`v=30n+2` satisfies the system of congruences, the largest of which is 242.