How to solve this very quickly? 
A number leaves a remainder $2$ when
  divided by $9$.Which of the following
  could not be the remainder when it is
  divided by $45$?



*

*$20$

*$30$

*$29$

*$38$


How could we solve this under a minute? Please explain your approach.

$x = 9k + 2$.
Now if $k = 5m + r$, $0 \leq r \leq 4$ then $x = 45m + 9r + 2$.
Thus the remainder upon dividing by $45$ should be of the form $9r + 2$. $30$ isn't it.
In other words, $x-2$ is divisible by $9$, hence the remainder upon dividing $x-2$ by $45$ should also be divisible by $9$, as $45$ is divisible by $9$.
In general, if $x$ leaves a remainder $r$ upon dividing by $a$, the possible remainders it leaves when divided by $b$ are of the form $k \times \text{gcd}(a,b) + r$. ($\text{gcd}$ = greatest common divisor).
In this case, $\text{gcd}(9,45) = 9$.
If $a$ and $b$ are co-prime, then any remainder is possible.
See: Bezout's Identity.
