N 1

Art of Problem Solving

AoPS Online

PEN Math problems from the collection "Problems in Elementary Number Theory"

Math problems from the collection "Problems in Elementary Number Theory"

3 M G

BBookmark VNew Topic kLocked

PEN Math problems from the collection "Problems in Elementary Number Theory"

Math problems from the collection "Problems in Elementary Number Theory"

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

N 1

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3615 posts

Peter

#1 h May 25, 2007, 12:25 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Show that the sequence $\{a_{n}\}_{n \ge 1}$ defined by $a_{n}=\lfloor n\sqrt{2}\rfloor$ contains an infinite number of integer powers of $2$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

337 posts

#2 h Oct 19, 2007, 7:22 PM • 5 Y

Y by Adventure10, Mango247, and 3 other users

Suppose the sequence contains finitely many powers of $2$ . Let $2^t$ the greatest power met in the sequence. Take $n>t$ . Then for some $m$ , $m\sqrt2<2^n<2^n+1<(m+1)\sqrt2$ , implying $2^n>m\sqrt2\ge2^n-(\sqrt2-1)>2^n-\dfrac12$ . Let $m\sqrt2=2^n-\epsilon$ . Then $\epsilon<\dfrac12$ . Let $k$ be a positive integer such that $2^k\epsilon<1<2^{k+1}\epsilon$ . Clearly $k\ge1$ and $\dfrac12<2^k\epsilon$ . This means $2^{n+k}>2^km\sqrt2=2^{n+k}-2^k\epsilon>2^{n+k}-1$ . We must also have $2^km\sqrt2=2^{n+k}-2^k\epsilon>2^{n+k}-(\sqrt2-1)>2^{n+k}-\dfrac12$ , which is $2^k\epsilon<\dfrac12$ , contradiction.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

153 posts

#3 h Mar 28, 2008, 3:29 AM • 2 Y

Y by Adventure10, Mango247

freemind wrote:

. We must also have $2^km\sqrt2 = 2^{n + k} - 2^k\epsilon > 2^{n + k} - (\sqrt2 - 1) > 2^{n + k} - \dfrac12$ , which is $2^k\epsilon < \dfrac12$ , contradiction.

Why is it true? Can you explain please?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1363 posts

Erken

#4 h May 28, 2008, 5:34 AM • 3 Y

Y by viperstrike, Adventure10, Mango247

Another solution:
Let us represent $\sqrt{2}$ in binary expression, $\sqrt{2}=a_{0}.a_{1}a_{2}\dots$ ,since $\sqrt{2}$ is irrational,we know that there are infinitely many $k$ ,such that $a_{k}=1$ .So $2^{k-1}\sqrt{2}=a_{0}a_{1}\dots a_{k-1}.a_{k}\dots$ .
Let $m=[2^{k-1}\sqrt{2}]$ ,where $[n]$ -means integer part of $n$ .Since $a_{k}=1$ ,it follows
$2^{k-1}\sqrt{2}-1<m<2^{k-1}\sqrt{2}-\frac{1}{2}$ ,hence
$2^{k}<(m+1)\sqrt{2}<2^{k}+\frac{1}{\sqrt{2}}$ ,and thus $\boxed{[(m+1)\sqrt{2}]=2^{k}}$ ,so we are done.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a