Difference between revisions of "2007 iTest Problems/Problem 47"
Clarkculus (talk | contribs) |
Clarkculus (talk | contribs) (→Solution) |
||
Line 12: | Line 12: | ||
== 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 11:46, 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 | |
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 |