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>
 
<math>2x_0 = a_1(a_2 a_3 \cdots)_2</math>
  

Revision as of 01:05, 3 January 2016

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 $b_1 = 0$ 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 $5$ 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 $2^5 - 1 = \boxed{31}$

-mathman523


$\fbox{D}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 30
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All AHSME Problems and Solutions

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