Difference between revisions of "2010 AIME I Problems/Problem 10"
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== Solution 2 == | == 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: | 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: | ||
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Our answer is thus <math>6 + 0 + 196 = \boxed{202}</math>. | 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*} | ||
+ | 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*} | ||
+ | </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 == | == See Also == |
Latest revision as of 07:07, 15 August 2024
Contents
Problem
Let be the number of ways to write in the form , where the 's are integers, and . An example of such a representation is . Find .
Solution 1
If we choose and such that there is a unique choice of and that makes the equality hold. So is just the number of combinations of and we can pick. If or we can let be anything from to . If then or . Thus .
Solution 2
Note that is the base representation of any number from to , and similarly is ten times the base representation of any number from to . Thus, the number of solutions is just the number of solutions to where , which is equal to as can range from to .
Solution 3
Note that and . It's easy to see that exactly 10 values in that satisfy our first congruence. Similarly, there are 10 possible values of for each choice of . Thus, there are possible choices for and . We next note that if and are chosen, then a valid value of determines , so we dive into some simple casework:
- If , there are 3 valid choices for . There are only 2 possible cases where , namely . Thus, there are possible representations in this case.
- If , can only equal 0. However, this case cannot occur, as . Thus, . However, . Thus, we have always.
- If , then there are 2 valid choices for . Since there are 100 possible choices for and , and we have already checked the other cases, it follows that choices of and fall under this case. Thus, there are possible representations in this case.
Our answer is thus .
Solution 4: Casework and Brute Force
We immediately see that can only be , or . We also note that the maximum possible value for is . We then split into cases:
Case 1: . We try to find possible values of . We plug in and to our initial equation, which gives us . Thus . We also see that . We now take these values of and find the number of pairs that work. If , . We see that there are possible pairs in this case. Using the same logic, there are ways for . For , we get the equation , for 2 ways. Thus, for , there are ways.
Case 2: . This case is almost identical to the one above, except . We also get 100 ways.
Case 3: . If , our initial equation becomes . It is obvious that , and we are left with . We saw above that there are ways.
Totaling everything, we get that there are ways.
Solution 5: Generating Functions
We will represent the problem using generating functions. Consider the generating function where the first factor represents , the second factor , and so forth. We want to find the coefficient of in the expansion of . Now rewriting each factor using the geometric series yields The coefficient of in this is simply , 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 has to be a single-digit number(, , or to be exact), and that has to be a two-digit multiple of ten. Then, , , and can be represented as follows: , where , , , , , and are all(not necessarily nonzero) digits. Now, we can write our given equation as follows: Now, each integer between and inclusive can be represented in exactly one way as , and this corresponds with one unique , so it remains to count the number of integers between and inclusive. This is easily counted to be .
Solution 6
Just note that this corresponds to , because we can use to fill in the remaining gap. Then, dividing by , we have , of which there are solutions.
Video Solution
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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