AIME 2010 I · 第 10 题

Solution 1

If we choose $a_3$ and $a_1$ such that $(10^3)(a_3) + (10)(a_1) \leq 2010$ there is a unique choice of $a_2$ and $a_0$ that makes the equality hold. So $N$ is just the number of combinations of $a_3$ and $a_1$ we can pick. If $a_3 = 0$ or $a_3 = 1$ we can let $a_1$ be anything from $0$ to $99$. If $a_3 = 2$ then $a_1 = 0$ or $a_1 = 1$. Thus $N = 100 + 100 + 2 = \fbox{202}$.

Solution 2

Note that $a_2\cdot 10^2 + a_0$ is the base $100$ representation of any number from $0$ to $9999$, and similarly $10(a_3\cdot 10^2 + a_1)$ is ten times the base $100$ representation of any number from $0$ to $9999$. Thus, the number of solutions is just the number of solutions to $2010 = 10a+b$ where $0\le a, b\le 9999$, which is equal to $\boxed{202}$ as $a$ can range from $0$ to $201$.

Solution 3

Note that $a_0 \equiv 2010\ (\textrm{mod}\ 10)$ and $a_1 \equiv 2010 - a_0\ (\textrm{mod}\ 100)$. It's easy to see that exactly 10 values in $0 \leq a_0 \leq 99$ that satisfy our first congruence. Similarly, there are 10 possible values of $a_1$ for each choice of $a_0$. Thus, there are $10 \times 10 = 100$ possible choices for $a_0$ and $a_1$. We next note that if $a_0$ and $a_1$ are chosen, then a valid value of $a_3$ determines $a_2$, so we dive into some simple casework:

• If $2010 - 10a_1 - a_0 \geq 2000$, there are 3 valid choices for $a_3$. There are only 2 possible cases where $2010 - 10a_1 - a_0 \geq 2000$, namely $(a_1, a_0) = (1,0), (10,0)$. Thus, there are $3 \times 2 = 6$ possible representations in this case.
• If $2010 - 10a_1 - a_0 < 1000$, $a_3$ can only equal 0. However, this case cannot occur, as $10a_1+a_0\leq 990+99 = 1089$. Thus, $2010-10a_1-a_0 \geq 921$. However, $2010-10a_1-a_0 = 1000a_3 + 100a_2 \equiv 0\ (\textrm{mod}\ 100)$. Thus, we have $2010-10a_1-a_0 \geq 1000$ always.
• If $1000 \leq 2010 - 10a_1 - a_0 < 2000$, then there are 2 valid choices for $a_3$. Since there are 100 possible choices for $a_0$ and $a_1$, and we have already checked the other cases, it follows that $100 - 2 - 0 = 98$ choices of $a_0$ and $a_1$ fall under this case. Thus, there are $2 \times 98 = 196$ possible representations in this case.

Our answer is thus $6 + 0 + 196 = \boxed{202}$.

Solution 4: Casework and Brute Force

We immediately see that $a_3$ can only be $0$, $1$ or $2$. We also note that the maximum possible value for $10a_1 + a_0$ is $990 + 99 = 1089$. We then split into cases:

Case 1: $a_3 = 0$. We try to find possible values of $a_2$. We plug in $a_3 = 0$ and $10a_1 + a_0 = 1089$ to our initial equation, which gives us $2010 = 0 + 100a_2 + 1089$. Thus $a_2 \geq 10$. We also see that $a_2 \leq 20$. We now take these values of $a_2$ and find the number of pairs $(a_1, a_0)$ that work. If $a_2 = 10$, $10a_1 + a_0 = 1010$. We see that there are $8$ possible pairs in this case. Using the same logic, there are $10$ ways for $a_2 = 11, 12 \ldots 19$. For $a_2 = 20$, we get the equation $10a_1 + a_0 = 10$, for 2 ways. Thus, for $a_3 = 0$, there are $8 + 10 \cdot 9 + 2 = 100$ ways.

Case 2: $a_3 = 1$. This case is almost identical to the one above, except $0 \leq a_2 \leq 10$. We also get 100 ways.

Case 3: $a_3 = 2$. If $a_3 = 2$, our initial equation becomes $100a_2 + 10a_1 + a_0 = 10$. It is obvious that $a_2 = 0$, and we are left with $10a_1 + a_0 = 10$. We saw above that there are $2$ ways.

Totaling everything, we get that there are $100 + 100 + 2 = \boxed{202}$ ways.

Solution 5: Generating Functions

We will represent the problem using generating functions. Consider the generating function

$$f(x) = (1+x^{1000}+x^{2000}+\cdots+x^{99000})(1+x^{100}+x^{200}+\cdots+x^{9900})(1+x^{10}+x^{20}+\cdots+x^{990})(1+x+x^2+\cdots+x^{99})$$ where the first factor represents $a_3$, the second factor $a_2$, and so forth. We want to find the coefficient of $x^{2010}$ in the expansion of $f(x)$. Now rewriting each factor using the geometric series yields

$$f(x) = \frac{\cancel{x^{100}-1}}{x-1} \cdot \frac{\cancel{x^{1000}-1}}{x^{10}-1} \cdot \frac{x^{10000}-1}{\cancel{x^{100}-1}} \cdot \frac{x^{100000}-1}{\cancel{x^{1000}-1}}=\frac{x^{10000}-1}{x-1} \cdot \frac{x^{100000}-1}{x^{10}-1} = (1+x+x^2+\cdots + x^{9999})(1+x^{10}+x^{20}+\cdots+x^{99990})$$ The coefficient of $x^{2010}$ in this is simply $\boxed{202}$, as we can choose any of the first 202 terms from the second factor and pair it with exactly one term in the first factor.

~rzlng

Solution 5

First note that $a_3$ has to be a single-digit number($0$, $1$, or $2$ to be exact), and that $a_1$ has to be a two-digit multiple of ten. Then, $a_3$, $a_2$, $a_1$ and $a_0$ can be represented as follows:

$$\begin{aligned} a_3 = a \\ a_2 = 10b+c \\ a_1= 10d+e \\ a_0 = 10f \end{aligned}$$ , where $a$, $b$, $c$, $d$, $e$, and $f$ are all(not necessarily nonzero) digits. Now, we can write our given equation as follows:

$$\begin{aligned} 2010 = 1000(a+b) + 100(c+d) + 10(e+f) \\ 201 = 100(a+b) + 10(c+d) + (e+f) \\ 201 = (100a + 10c + e) + (100b + d + f) \end{aligned}$$ Now, each integer between $0$ and $201$ inclusive can be represented in exactly one way as $100a + 10c + e$, and this corresponds with one unique $100b + d + f$, so it remains to count the number of integers between $0$ and $201$ inclusive. This is easily counted to be $\boxed{202}$.

Solution 6

Just note that this corresponds to $0 \leq 10^3 \cdot a_3 + 10^2 \cdot a_2 + 10^1 \cdot a_1 \leq 2010$, because we can use $a_0$ to fill in the remaining gap. Then, dividing by $10$, we have $0 \leq \overline{a_3 a_2 a_1} \leq 201$, of which there are $\boxed{202}$ solutions.

Video Solution

https://youtu.be/DNC2zd5rReY