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 solution is just the number of solutions to <math>2010 = 10a+b</math> where <math>0\le a, b\le 9999</math>, which is clearly equal to <math>\boxed{202}</math> as <math>a</math> can range from <math>0</math> to <math>201</math>. | ||
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+ | == 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 | + | ==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: | 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: | ||
Revision as of 20:24, 15 January 2018
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 solution is just the number of solutions to where , which is clearly 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.
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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