How many ways to write $2010$? 
Let $ N$ be the number of ways to write $ 2010$ in the form $ 2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$, where the $ a_i$'s are integers, and $ 0 \le a_i \le 99$. An example of such a representation is $ 1\cdot10^3 + 3\cdot10^2 + 67\cdot10^1 + 40\cdot10^0$. Find $ N$.

I picked the biggest $a_1$ so: $a_1 = 2$, there are only two ways to form $2010$. 
Take $a_1 = 1$ now. This opens up to a lot of possibilities.
Specific Casework should work:
Cases 1-1: $a_2 = 10$, then possibilities are: $a_1 = 1, a_0 = 0$ or $a_1  = 0, a_0 = 10$
Actually, I think a number-theoretic way is easier.
But still. 
Case 1: $a_1 = 1$ then we must solve: 
$100x + 10y + z = 1010$. 
Since $0 \le x \le 10$, we can casework $x$ so that:
Case 1-1:$x = 0$. So that:
$10y + z = 1010 \implies z \equiv 0 \pmod{10}, z = 10k$ and $y = 101 - k$. 
Hence, $(0, 101 - k, 10k)$. $\min{k} = 0 $ and we need to find the max of $k$. We must have, $101 - k \le 99$ and $10k \le 99$. This suggests, $k \le 9$.
Cases 1-2: $x=1$. So that:
$10y + z = 910 \implies z \equiv 0 \pmod{10}$ Again, $z = 10k$ and $y = 91 - k$. Giving a set of $(1, 91 - k, 10k)$.Here again, $\min{k} = 0$ and $10k \le 99$ so $k \le 9$. 
I am conjecturing that since we are always increasing $x$ values, the value on the RHS will always be divisible by $10$. 
$x = 9$ so that:
$10y + z = 110 \implies z = 10k$ and $y = 11 - k$, which again there are $9$ values.
Except if $x=10$ then there is: $10y + z = 10$ then $z = 10k$ and $y = 1 - k$. Then $k$ must be $1$. 
So there are: $10(9) + 1 + 2 = 93$ solutions total.
This is just an attempt!
Bump: anybody have anything?
 A: The generating function for these representations is
$$
\frac{x^{100\cdot1000}-1}{x^{1000}-1}\cdot\frac{x^{100\cdot100}-1}{x^{100}-1}\cdot\frac{x^{100\cdot10}-1}{x^{10}-1}\cdot\frac{x^{100}-1}{x-1}=\frac{x^{100000}-1}{x^{10}-1}\cdot\frac{x^{10000}-1}{x-1}\;,
$$
which is also the generating function for representations of the form $b_1\cdot10+b_0$ with $0\le b_i\le9999$. Since this latter restriction is irrelevant for representations of $2010$, we are simply looking for the number of ways to represent $2010$ by tens and ones. We can have anything from $0$ to $201$ tens, so the number of such representations is $202$.
In response to Calvin Lin's answer: The cancellation in the generating functions corresponds to a bijection between the two forms of representation, with $b_1=100a_3+a_1$ and $b_0=100a_2+a_0$. 
A: It's easier to start from the other end.  We observe first that $a_0$ can be any multiple of $10$ from $0$ through $90$.  There are, of course, ten such values.
Then $a_1$ can be any one- or two-digit value equivalent to $11-(a_0/10) \bmod 10$.  Again, there are ten such values.
The choices of $a_2$ are much more limited by the fact that the target sum is only $2010$.  In which cases are there two different values?  In which cases are there three?
Once $a_0, a_1, a_2$ have been decided, there is only one usable value of $a_3$.
A: Here's the bijection approach, which is hinted at by Joriki's solution.
Let $a_i = 10 b_i + c_i$, where $ 0 \leq c_i \leq 9$, $ 0 \leq b_i \leq 9$.
Then, we have   
$$2010 = 10 (1000b_3 + 100 b_2 + 10b_1 + b_0) + ( 1000 c_3 + 100 c_2 + 10 c_1 + c_0)$$
Thus, given any representation of $2010 = 10 B + C $, there is a unique corresponding $b_i, c_i$ (clearly the digits of B and C) that we can create, and vice versa. This establishes the bijection. Hence, the answer is 202.
A: Here's a solution that's not particularly elegant but gives the right answer (checked by brute force).
First express each coefficient uniquely as $b_i\cdot 10 + a_i$.
$2010 = b_3 \cdot 10^4 + a_3 \cdot 10^3 + b_2 \cdot 10^3 + a_2 \cdot 10^2 +
b_1 \cdot 10^2 + a_1 \cdot 10 + b_0 \cdot 10 + a_0$.
We must have $b_3 = 0, a_0 = 0$, so
$201 = (a_3 + b_2) \cdot 10^2 + (a_2 + b_1) \cdot 10^1 + (a_1 + b_0)$.
Each coefficient is in $0, \ldots, 18$.
If $a_3 + b_2 = 1$, then $101 = (a_2 + b_1) \cdot 10^1 + (a_1 + b_0)$. We have
$a_2 + b_1 = 9, a_1 + b_0 = 11$ or $a_2 + b_1 = 10, a_1 + b_0 = 1$, this gives
$2 \cdot (10 \cdot 8 + 9 \cdot 2) = 196$ solutions.
If $a_3 + b_2 = 2$, then $1 = (a_2 + b_1) \cdot 10^1 + (a_1 + b_0)$. We have
$a_2 + b_1 = 0, a_1 + b_0 = 1$, this gives $3 \cdot 2 = 6$ solutions.
So together we have 202 solutions.
