A little use of number theory
Suppose that you have $1995$ different integers written on a paper sheet, taken at a random.Can you put them along a circle so that for every two adjacent numbers,the ratio of the larger and smaller is a prime,not necessarily distinct,I think
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
Re: A little use of number theory
No.Actually it is not possible for odd numbers$\ge 1$.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules
Re: A little use of number theory
My solution:
Actually it's impossible for any 1995 integers.Suppose it's possible.Then let $a_{0},a_{1},....,a_{1995}$ be the integers in clockwise direction.Then $a_{k-1}/a_{k}$ is a prime or the reciprocal of a prime for $k=1,2,.....,1995$ with $a_{0}=a_{1995}$. Suppose, $m$ of these are reciprocals of primes.Since
\[\frac{a_0}{a_1}\cdot\frac{a_1}{a_2}\cdots\frac{a_{1994}}{a_{1995}}=1\]
This means the product of $m$ primes is equal to the product of $1995-m$ primes.Unique prime factorization shows that $1995-m=m$,which is impossible(why?) and brings the contradiction.
Actually it's impossible for any 1995 integers.Suppose it's possible.Then let $a_{0},a_{1},....,a_{1995}$ be the integers in clockwise direction.Then $a_{k-1}/a_{k}$ is a prime or the reciprocal of a prime for $k=1,2,.....,1995$ with $a_{0}=a_{1995}$. Suppose, $m$ of these are reciprocals of primes.Since
\[\frac{a_0}{a_1}\cdot\frac{a_1}{a_2}\cdots\frac{a_{1994}}{a_{1995}}=1\]
This means the product of $m$ primes is equal to the product of $1995-m$ primes.Unique prime factorization shows that $1995-m=m$,which is impossible(why?) and brings the contradiction.
Last edited by Masum on Sat Oct 13, 2012 8:38 pm, edited 1 time in total.
Reason: Properly Latexed. It is safe to use the preview button. :)
Reason: Properly Latexed. It is safe to use the preview button. :)
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
- nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
- Location:24.758613,90.400161
- Contact:
Re: A little use of number theory
Sanzeed edit it.
i would give a pm.but, this request is also to the moderators
i would give a pm.but, this request is also to the moderators
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
Re: A little use of number theory
My solve has only three lines.I've given first two in hint.It's your task to find the third.Again my solve works for every odd number$\ge 3$ to show that's impossible.SANZEED wrote:My solution:
Actually it's impossible for any 1995 integers.Suppose it's possible.Then let $a_{0},a_{1},....,a_{1995}$ be the integers in clockwise direction.Then $a_{k-1}/a_{k}$ is a prime or the reciprocal of a prime for $k=1,2,.....,1995$ with $a_{0}=a_{1995}$.Suppose $m$ of these are reciprocals of primes.Since
$(a_{0}/a_{1})(a_{1}/a_{2}).....(a_{1994}/a_{1995})=1$
This means the product of $m$ primes is equal to the product of $1995-m$ primes.Unique prime factorization shows that $1995-m=m$,which is impossible(why?) and brings the contradiction.
Hint :
Welcome to BdMO Online Forum. Check out Forum Guides & Rules