Difference between revisions of "2010 AIME I Problems/Problem 10"

Latest revision as of 07:07, 15 August 2024

Problem

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 $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ .

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: $\begin{align*} a_3 = a \\ a_2 = 10b+c \\ a_1= 10d+e \\ a_0 = 10f \end{align*}$ , where $a$ , $b$ , $c$ , $d$ , $e$ , and $f$ are all(not necessarily nonzero) digits. Now, we can write our given equation as follows: $\begin{align*} 2010 = 1000(a+b) + 100(c+d) + 10(e+f) \\ 201 = 100(a+b) + 10(c+d) + (e+f) \\ 201 = (100a + 10c + e) + (100b + d + f) \end{align*}$ 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

@@ Line 2: / Line 2: @@
 Let <math>N</math> be the number of ways to write <math>2010</math> in the form <math>2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0</math>, where the <math>a_i</math>'s are integers, and <math>0 \le a_i \le 99</math>. An example of such a representation is <math>1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0</math>. Find <math>N</math>.
-== Solution ==
+== Solution 1 ==
 If we choose <math>a_3</math> and <math>a_1</math> such that <math>(10^3)(a_3) + (10)(a_1) \leq 2010</math> there is a [[bijection|unique]] choice of <math>a_2</math> and <math>a_0</math> that makes the equality hold.  So <math>N</math> is just the number of combinations of <math>a_3</math> and <math>a_1</math> we can pick.  If <math>a_3 = 0</math> or <math>a_3 = 1</math> we can let <math>a_1</math> be anything from <math>0</math> to <math>99</math>.  If <math>a_3 = 2</math> then <math>a_1 = 0</math> or <math>a_1 = 1</math>.  Thus <math>N = 100 + 100 + 2 = \fbox{202}</math>.
-== See also ==
+== Solution 2 ==
+Note that <math>a_2\cdot 10^2 + a_0</math> is the base <math>100</math> representation of any number from <math>0</math> to <math>9999</math>, and similarly <math>10(a_3\cdot 10^2 + a_1)</math> is ten times the base <math>100</math> representation of any number from <math>0</math> to <math>9999</math>. Thus, the number of solutions is just the number of solutions to <math>2010 = 10a+b</math> where <math>0\le a, b\le 9999</math>, which is equal to <math>\boxed{202}</math> as <math>a</math> can range from <math>0</math> to <math>201</math>.
+== Solution 3 ==
+Note that <math>a_0 \equiv 2010\ (\textrm{mod}\ 10)</math> and <math>a_1 \equiv 2010 - a_0\  (\textrm{mod}\ 100)</math>. It's easy to see that exactly 10 values in <math>0 \leq a_0 \leq 99</math> that satisfy our first congruence. Similarly, there are 10 possible values of <math>a_1</math> for each choice of <math>a_0</math>. Thus, there are <math>10 \times 10 = 100</math> possible choices for <math>a_0</math> and <math>a_1</math>. We next note that if <math>a_0</math> and <math>a_1</math> are chosen, then a valid value of <math>a_3</math> determines <math>a_2</math>, so we dive into some simple casework:
+*If <math>2010 - 10a_1 - a_0 \geq 2000</math>, there are 3 valid choices for <math>a_3</math>. There are only 2 possible cases where <math>2010 - 10a_1 - a_0 \geq 2000</math>, namely <math>(a_1, a_0) = (1,0), (10,0)</math>. Thus, there are <math>3 \times 2 = 6</math> possible representations in this case.
+*If <math>2010 - 10a_1 - a_0 < 1000</math>, <math>a_3</math> can only equal 0. However, this case cannot occur, as <math>10a_1+a_0\leq 990+99 = 1089</math>. Thus, <math>2010-10a_1-a_0 \geq 921</math>. However,  <math>2010-10a_1-a_0 = 1000a_3 + 100a_2 \equiv 0\  (\textrm{mod}\ 100)</math>. Thus, we have <math>2010-10a_1-a_0 \geq 1000</math> always.
+*If <math>1000 \leq 2010 - 10a_1 - a_0 < 2000</math>, then there are 2 valid choices for <math>a_3</math>. Since there are 100 possible choices for <math>a_0</math> and <math>a_1</math>, and we have already checked the other cases, it follows that <math>100 - 2 - 0 = 98</math> choices of <math>a_0</math> and <math>a_1</math> fall under this case. Thus, there are <math>2 \times 98 = 196</math> possible representations in this case.
+Our answer is thus <math>6 + 0 + 196 = \boxed{202}</math>.
+==Solution 4: Casework and Brute Force==
+We immediately see that <math>a_3</math> can only be <math>0</math>, <math>1</math> or <math>2</math>. We also note that the maximum possible value for <math>10a_1 + a_0</math> is <math>990 + 99 = 1089</math>. We then split into cases:
+Case 1: <math>a_3 = 0</math>.
+We try to find possible values of <math>a_2</math>. We plug in <math>a_3 = 0</math> and <math>10a_1 + a_0 = 1089</math> to our initial equation, which gives us <math>2010 = 0 + 100a_2 + 1089</math>. Thus <math>a_2 \geq 10</math>. We also see that <math>a_2 \leq 20</math>. We now take these values of <math>a_2</math> and find the number of pairs <math>(a_1, a_0)</math> that work. If <math>a_2 = 10</math>, <math>10a_1 + a_0 = 1010</math>. We see that there are <math>8</math> possible pairs in this case. Using the same logic, there are <math>10</math> ways for <math>a_2 = 11, 12 \ldots 19</math>. For <math>a_2 = 20</math>, we get the equation <math>10a_1 + a_0 = 10</math>, for 2 ways. Thus, for <math>a_3 = 0</math>, there are <math>8 + 10 \cdot 9 + 2 = 100</math> ways.
+Case 2: <math>a_3 = 1</math>.
+This case is almost identical to the one above, except <math>0 \leq a_2 \leq 10</math>. We also get 100 ways.
+Case 3: <math>a_3 = 2</math>.
+If <math>a_3 = 2</math>, our initial equation becomes <math>100a_2 + 10a_1 + a_0 = 10</math>. It is obvious that <math>a_2 = 0</math>, and we are left with <math>10a_1 + a_0 = 10</math>. We saw above that there are <math>2</math> ways.
+Totaling everything, we get that there are <math>100 + 100 + 2 = \boxed{202}</math> ways.
+==Solution 5: Generating Functions==
+We will represent the problem using generating functions. Consider the generating function <cmath>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})</cmath>
+where the first factor represents <math>a_3</math>, the second factor <math>a_2</math>, and so forth. We want to find the coefficient of <math>x^{2010}</math> in the expansion of <math>f(x)</math>. Now rewriting each factor using the geometric series yields <cmath>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})</cmath>
+The coefficient of <math>x^{2010}</math> in this is simply <math>\boxed{202}</math>, 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 <math>a_3</math> has to be a single-digit number(<math>0</math>, <math>1</math>, or <math>2</math> to be exact), and that <math>a_1</math> has to be a two-digit multiple of ten.
+Then, <math>a_3</math>, <math>a_2</math>, <math>a_1</math> and <math>a_0</math> can be represented as follows:
+<cmath>\begin{align*}
+a_3 = a
+\\ a_2 = 10b+c
+\\ a_1= 10d+e
+\\ a_0 = 10f
+\end{align*}</cmath>
+, where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math> are all(not necessarily nonzero) digits.
+Now, we can write our given equation as follows:
+<cmath>
+\begin{align*}
+= 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{align*}
+</cmath>
+Now, each integer between <math>0</math> and <math>201</math> inclusive can be represented in exactly one way as <math>100a + 10c + e</math>, and this corresponds with one unique <math>100b + d + f</math>, so it remains to count the number of integers between <math>0</math> and <math>201</math> inclusive. This is easily counted to be <math>\boxed{202}</math>.
+==Solution 6 ==
+Just note that this corresponds to <math>0 \leq 10^3 \cdot a_3 + 10^2 \cdot a_2 + 10^1 \cdot a_1 \leq 2010</math>, because we can use <math>a_0</math> to fill in the remaining gap. Then, dividing by <math>10</math>, we have <math>0 \leq \overline{a_3 a_2 a_1} \leq 201</math>, of which there are <math>\boxed{202}</math> solutions.
+==Video Solution==
+https://youtu.be/DNC2zd5rReY
+== See Also ==
 {{AIME box|year=2010|num-b=9|num-a=11|n=I}}
 [[Category:Intermediate Number Theory Problems]]
+{{MAA Notice}}

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