Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1993 AHSME Problems/Problem 30"

(→‎Solution)
(→‎Solution 2)
 
(19 intermediate revisions by 4 users not shown)
Line 18: Line 18:
 
We are going to look at this problem in binary.  
 
We are going to look at this problem in binary.  
  
−
<cmath>x_0 = (0.a_1 a_2 \cdots )_2</cmath>
+
<math>x_0 = (0.a_1 a_2 \cdots )_2</math>
  
 +
<math>2x_0 = (a_1.a_2 a_3 \cdots)_2</math>
  
−
<cmath>2x_0 = a_1(a_2 a_3 \cdots)_2</cmath>
+
If <math>2x_0 < 1</math>, then <math>x_0 < \frac{1}{2}</math> which means that <math>a_1 = 0</math> and so <math>x_1 = (.a_2 a_3 a_4 \cdots)_2</math>
−
 
 
−
If <math>2x_0 < 1</math>, then <math>x_0 < \frac{1}{2}</math> which means that <math>b_1 = 0</math> and so <math>x_1 = (.a_2 a_3 a_4 \cdots)_2</math>
 
  
 
If <math>2x_0 \geq 1</math> then <math>x \geq \frac{1}{2}</math> which means that <math>x_1 = 2x_0 - 1 = (.a_2 a_3 a_4 \cdots)_2</math>.
 
If <math>2x_0 \geq 1</math> then <math>x \geq \frac{1}{2}</math> which means that <math>x_1 = 2x_0 - 1 = (.a_2 a_3 a_4 \cdots)_2</math>.
Line 29: Line 28:
 
Using the same logic, we notice that this sequence cycles and that since <math>x_0 = x_5</math> we notice that <math>a_n = a_{n+5}</math>.  
 
Using the same logic, we notice that this sequence cycles and that since <math>x_0 = x_5</math> we notice that <math>a_n = a_{n+5}</math>.  
  
−
We have <math>5</math> possibilities for each of <math>a_1</math> to <math>a_5</math> but we can't have <math>a_1 = a_2 = a_3 = a_4 = a_5 = 1</math> so we have <math>2^5 - 1 = \boxed{31}</math>
+
We have <math>2</math> possibilities for each of <math>a_1</math> to <math>a_5</math> but we can't have <math>a_1 = a_2 = a_3 = a_4 = a_5 = 1</math> so we have <math>2^5 - 1 = \boxed{(D)31}</math>
  
 
-mathman523
 
-mathman523
  
 +
== Solution 2==
 +
 +
<math>x_5 = \left\{ 2^5x_0 \right\}</math>
 +
 +
<math>\therefore x_0 = \left\{ 2^5x_0 \right\}</math>
 +
 +
<math>Suppose \  x_0 = 2^5 x_0 - k \ (k \geq 0)</math>
 +
 +
<math>31x_0 = k</math>
 +
 +
<math>x_0 = \frac{k}{31}</math>
 +
 +
<math>\therefore k = 0, 1, 2,..., 30</math>
 +
 +
select (D)
  
−
<math>\fbox{D}</math>
+
~ JiYang
  
 
== See also ==
 
== See also ==

Latest revision as of 22:57, 24 October 2024

Problem

Given $0\le x_0<1$, let \[x_n=\left\{ \begin{array}{ll} 2x_{n-1} &\text{ if }2x_{n-1}<1 \\ 2x_{n-1}-1 &\text{ if }2x_{n-1}\ge 1 \end{array}\right.\] for all integers $n>0$. For how many $x_0$ is it true that $x_0=x_5$?

$\text{(A) 0} \quad \text{(B) 1} \quad \text{(C) 5} \quad \text{(D) 31} \quad \text{(E) }\infty$

Solution

We are going to look at this problem in binary.

$x_0 = (0.a_1 a_2 \cdots )_2$

$2x_0 = (a_1.a_2 a_3 \cdots)_2$

If $2x_0 < 1$, then $x_0 < \frac{1}{2}$ which means that $a_1 = 0$ and so $x_1 = (.a_2 a_3 a_4 \cdots)_2$

If $2x_0 \geq 1$ then $x \geq \frac{1}{2}$ which means that $x_1 = 2x_0 - 1 = (.a_2 a_3 a_4 \cdots)_2$.

Using the same logic, we notice that this sequence cycles and that since $x_0 = x_5$ we notice that $a_n = a_{n+5}$.

We have $2$ possibilities for each of $a_1$ to $a_5$ but we can't have $a_1 = a_2 = a_3 = a_4 = a_5 = 1$ so we have $2^5 - 1 = \boxed{(D)31}$

-mathman523

Solution 2

$x_5 = \left\{ 2^5x_0 \right\}$

$\therefore x_0 = \left\{ 2^5x_0 \right\}$

$Suppose \ x_0 = 2^5 x_0 - k \ (k \geq 0)$

$31x_0 = k$

$x_0 = \frac{k}{31}$

$\therefore k = 0, 1, 2,..., 30$

select (D)

~ JiYang

See also

1993 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 30
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_30&oldid=230599"