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Difference between revisions of "1993 AHSME Problems/Problem 30"
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<math>x_0 = (0.a_1 a_2 \cdots )_2</math>
 
<math>x_0 = (0.a_1 a_2 \cdots )_2</math>
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<math>2x_0 = a_1(a_2 a_3 \cdots)_2</math>
 
  
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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>
+
<math>2x_0 = (a_1.a_2 a_3 \cdots)_2</math>
 +
 
 +
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 \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 27: 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
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−
 
−
<math>\fbox{D}</math>
 
  
 
== See also ==
 
== See also ==

Revision as of 20:22, 1 June 2020

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

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

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