how many integers satisfy for modular aritmetic

Question

How many integers $n$ are there which satisfy $1\leq n \leq 2014$ and $21n = 25 \pmod {29}$?

Answer 1

Hint:

$$21^{-1} \equiv 18 \pmod{29}$$

Therefore:

$$18·21 n \equiv 18·25 \pmod{29} \Longrightarrow n \equiv 15 \pmod{29}$$

Answer 2

Hint: instead of using standard methods, this congruence is very easy to solve by simplifying the coefficients and cancelling: $$21n\equiv25\pmod{29}\quad\Leftrightarrow\quad -8n\equiv-4\pmod{29}\quad\Leftrightarrow\quad 2n\equiv1\pmod{29}$$ and so on.