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Difference between revisions of "2007 iTest Problems/Problem 47"
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== Solution ==
 
== Solution ==
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1447
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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
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<cmath>\begin{cases}
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1=\lambda_1+\lambda_2\\
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1=-\lambda_1+2\lambda_2
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\end{cases}</cmath>
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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>.
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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>.
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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>.
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~clarkculus
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==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 $\{X_n\}$ and $\{Y_n\}$ be sequences defined as follows: $X_0=Y_0=X_1=Y_1=1$,

\begin{align*}X_{n+1}&=X_n+2X_{n-1}\qquad(n=1,2,3\ldots),\\ Y_{n+1}&=3Y_n+4Y_{n-1}\qquad(n=1,2,3\ldots).\end{align*}

Let $k$ be the largest integer that satisfies all of the following conditions: $|X_i-k|\leq 2007$, for some positive integer $i$; $|Y_j-k|\leq 2007$, for some positive integer $j$; and $k<10^{2007}$. Find the remainder when $k$ is divided by $2007$.

Solution

First, we solve these linear recurrences. The characteristic polynomial of $X_n$ is $x^2=x+2$ which has roots -1 and 2. Then with the given values, $X_n=\lambda_1(-1)^n+\lambda_2(2)^n$ where $\lambda_1$ and $\lambda_2$ are the solutions to the system \[\begin{cases} 1=\lambda_1+\lambda_2\\ 1=-\lambda_1+2\lambda_2 \end{cases}\] Solving, we find $\lambda_1=1/3$ and $\lambda_2=2/3$, so $X_n=\frac{(-1)^n+2^{n+1}}{3}$. Similarly, $Y_n=\frac{3(-1)^n+2^{2n+1}}{5}$.

We can ignore the $(-1)^n$ terms because they will be inconsequential compared to the $2^n$ terms, so define $x_n=\frac{2^{n+1}}{3}$ and $y_n=\frac{2^{2n+1}}{5}$. Note that for some $a$ and $b$, $|X_a-Y_b|\le4014$ so that we can place $k$ at the average of $X_a$ and $Y_b$ at least. Therefore $X_a$ and $Y_b$ will have to be somewhat close to each other, so we examine the equation $X_a\approx x_a=Y_b\approx y_b$, or $\frac{2^a}{3}=\frac{2^{2b}}{5}$. Solving for $a$ results in $a=2b-\log_2(5/3)$, and because $a$ and $b$ are integers, $a=2b-1$.

Using our approximations in our inequality, we find $|\frac{2^{2b}}{3}-\frac{2^{2b+1}}{5}|\le4014$, which simplifies to $2^{2b}\le4014\cdot15$. Bashing or calculator use results in $b<8$, so $b=7$ and $a=13$. Note that $X_{13}<Y_7$, so the largest $k$ possible will be $X_{13}+2007=7468$. The requested answer is $7468\text{ mod }{2007}=\boxed{1447}$.

~clarkculus

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 46 		Followed by:
Problem 48
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
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