<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] 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 $a_3$.
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