<s>[INST] 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?
 [/INST] 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.
</s>