modular arithmetic with different mods When I am given that $N= 4\mod 5$ and that $N = 0\mod 9$, can I find what the remainder is when $N$ is divided by 45? I know very little of mods and only those that have the same $\mod m$, not of different ones, so I really don't know what to do. Here's a problem if you need to refer to anything...
Let $N=123456789101112\dots4344$ be the $79$-digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$?
 A: It might help your intuition to do a couple of these type of problems "by hand", without looking at modular inverses etc, just to get a feel for what is going on, before learning the techniques given in Chinese Remainder theorem discussions.
So looking at some small numbers $a$ where $a\equiv 4 \bmod 5$:
$a=4,9,14,19,24,29,34,39,44,49,54,59,\ldots$
and some numbers $b$ where $b\equiv 0 \bmod 9$:
$b=0,9,18,27,36,45,54,63,\ldots$
You can see the two matches in the lists of $9$ and $54$, and it shouldn't be surprising that they are $45$ apart, since the remainders of any number with respect to $5$ and $9$ will not be changed by adding a number that is a multiple of both $5$ and $9$. Since $45$ is the least common multiple of $5$ and $9$, this is the modulus to resolve into. And there will be exactly one common value less than this (if there is a solution at all, which there always will be when the original moduluses are coprime). This is fairly easy to satisfy yourself of by trying a few different cases in a small example like this.
