Difference between revisions of "2007 iTest Problems/Problem 47"
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== Solution == | == Solution == | ||
− | 1447 | + | First, we solve these linear recurrences. The characteristic polynomial of <math>X_n</math> is <math>x^2=x+2</math> which has roots -1 and 2. Then with the given values, <math>X_n=\lambda_1(-1)^n+\lambda_2(2)^n</math> where <math>\lambda_1</math> and <math>\lambda_2</math> are the solutions to the system |
+ | <cmath>\begin{cases} | ||
+ | 1=\lambda_1+\lambda_2\ | ||
+ | 1=-\lambda_1+2\lambda_2 | ||
+ | \end{cases}</cmath> | ||
+ | Solving, we find <math>\lambda_1=1/3</math> and <math>\lambda_2=2/3</math>, so <math>X_n=\frac{(-1)^n+2^{n+1}}{3}</math>. Similarly, <math>Y_n=\frac{3(-1)^n+2^{2n+1}}{5}</math>. | ||
+ | |||
+ | We can ignore the <math>(-1)^n</math> terms because they will be inconsequential compared to the <math>2^n</math> terms, so define <math>x_n=\frac{2^{n+1}}{3}</math> and <math>y_n=\frac{2^{2n+1}}{5}</math>. Note that for some <math>a</math> and <math>b</math>, <math>|X_a-Y_b|\le4014</math> so that we can place <math>k</math> at the average of <math>X_a</math> and <math>Y_b</math> at least. Therefore <math>X_a</math> and <math>Y_b</math> will have to be somewhat close to each other, so we examine the equation <math>X_a\approx x_a=Y_b\approx y_b</math>, or <math>\frac{2^a}{3}=\frac{2^{2b}}{5}</math>. Solving for <math>a</math> results in <math>a=2b-\log_2(5/3)</math>, and because <math>a</math> and <math>b</math> are integers, <math>a=2b-1</math>. | ||
+ | |||
+ | Using our approximations in our inequality, we find <math>|\frac{2^{2b}}{3}-\frac{2^{2b+1}}{5}|\le4014</math>, which simplifies to <math>2^{2b}\le4014\cdot15</math>. Bashing or calculator use results in <math>b<8</math>, so <math>b=7</math> and <math>a=13</math>. Note that <math>X_{13}<Y_7</math>, so the largest <math>k</math> possible will be <math>X_{13}+2007=7468</math>. The requested answer is <math>7468\text{ mod }{2007}=\boxed{1447}</math>. | ||
+ | |||
+ | ~clarkculus | ||
+ | |||
==See Also== | ==See Also== | ||
{{iTest box|year=2007|num-b=46|num-a=48}} | {{iTest box|year=2007|num-b=46|num-a=48}} |
Latest revision as of 10:46, 18 April 2024
Problem
Let and be sequences defined as follows: ,
Let be the largest integer that satisfies all of the following conditions: , for some positive integer ; , for some positive integer ; and . Find the remainder when is divided by .
Solution
First, we solve these linear recurrences. The characteristic polynomial of is which has roots -1 and 2. Then with the given values, where and are the solutions to the system Solving, we find and , so . Similarly, .
We can ignore the terms because they will be inconsequential compared to the terms, so define and . Note that for some and , so that we can place at the average of and at least. Therefore and will have to be somewhat close to each other, so we examine the equation , or . Solving for results in , and because and are integers, .
Using our approximations in our inequality, we find , which simplifies to . Bashing or calculator use results in , so and . Note that , so the largest possible will be . The requested answer is .
~clarkculus
See Also
2007 iTest (Problems, Answer Key) | ||
Preceded by: Problem 46 |
Followed by: Problem 48 | |
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