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1993 AHSME Problems/Problem 30: Difference between revisions

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<math>\therefore x_0 = \left\{ 2^5x_0 \right\}</math>
<math>\therefore x_0 = \left\{ 2^5x_0 \right\}</math>


Suppose <math>x_0 = 2^5x_0 - k (k \gep 0)</math>
Suppose <math>x_0 = 2^5 x_0 - k (k \gep 0)</math>


<math>31x_0 = k</math>
<math>31x_0 = k</math>

Revision as of 22:50, 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 $x_0 = 2^5 x_0 - k (k \gep 0)$ (Error compiling LaTeX. Unknown error_msg)

$31x_0 = k$

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

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

$\therefore 31$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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