Find a positive integer $a$ such that $4 \mid a$ and $9 \mid a+1$ and $25 \mid a+2$ I need to find a positive integer $a$ such that $4 \mid a$ and $9 \mid a+1$ and $25 \mid a+2$.
I tried converting this to a system of congruences, and I got the following.
$a \equiv 0$ mod $4$
$(a+1) \equiv 0$ mod $9$
$(a+2) \equiv 0$ mod $25$ 
And then I want to use Chinese Remainder Theorem. But, I think my congruences are wrong since they don't involve the variable $x$, but I can't imagine how it would fit in.  It seems unnecessary but I think I am supposed to use this method.
What should my congruences look like?

$a \equiv 0 \pmod{4}$, so $a = 4k$, and $a+1 \equiv 0 \pmod{9}$, it means $4k+1 \equiv 0 \pmod{9}$, the minimum $k$(here $k \ge 0$) is $k = 2$. so, when $a \equiv 0 \pmod{4}$ and $a+1 \equiv 0 \pmod{9}$, the minimum $a$ is $a = 4k = 8$, all the $a$ is $8+(4 \cdot 9)m = 8+36m$.
then, we have $a+2 \equiv 0 \pmod{25}$, it means $8+36m+2 \equiv 0 \pmod{25}$, as $36m \equiv 15 \pmod{25}$. the minimum $m$(here $m \ge 0$) is $m = 15$.
so $a = 8+36m = 548$ is one solution.
