I think it is 11. Clearly powers of 2 can be written since $2^k=\frac{2^{k+2}-2^{k+1}}{2^2-2}$ and of course, if $x$ can be written, so does $2^px$ for all $p$. Next, 1 can be written, so does $3=\frac{2^3-2}{2^2-2}$; $5=\frac{2^5-2}{2^3-2}$, $7=\frac{2^4-2}{2^2-2}$ and $9=\frac{2^7-2}{2^4-2}$. Suppose that we can write 11 like that and take $a>b$. Then $c>d$ and $ 2^b(2^{a-b}-1)=11\cdot 2^d\cdot (2^{c-d}-1)$. Thus $b=d$ and since $11$ divides $2^{a-b}-1$ we must have $a-b$ multiple of 10. Write the equation as $ 2^{a-b}+10=11\cdot 2^{c-d}$ and deduce that $ c-d=1$ since otherwise 4 divides 10. But then $a-b$ is not a multiple of 10, false.

This problem is not hard,you can prove that $1=2^2-2/2^2-2,2=2^3-2^2/2^2-2..............,10=2^6-2^2/2^3-2$,but $11$is not$= \frac{2^{a} - 2^{b}}{2^{c} - 2^{d}}$

Well you cant just reason by trial and error. How can you be sure that you have exausted EACH AND EVERY possiblity for the required $a,b,c,d$ ???? A mathematical proof needs to be water tight. I must say, your solution has TOO many loopholes . Please post a proper solution if you have one!

Let's looking at this problem. Prove that in the interval$(2^{p+1}-1;2^p+1)$,wher $p$ is a prime numbre,don't existe odd numbre that be represented in the form $\frac{2^{a} - 2^{b}}{2^{c} - 2^{d}}$

hey its easy to prove that nos. of the form 2^k +1 and 2^k -1 can be represented in this form. further we can prove that 2^s(2^k +1) and 2^s(2^k -1) can also be represented in this form. then we can prove that no other nos. can be represented in this form. its easy to note that 11 is the 1st no that cannot be represented in the form 2^k -1 & 2^k +1. this is how i solved it.hope it works ur purpose.

venkata wrote: then we can prove that no other nos. can be represented in this form Please use LaTeX, and please give your proof.

Rushil wrote: least positive integer that cannot be represented in the form $\frac{2^{a} - 2^{b}}{2^{c} - 2^{d}}$ A representable number is of the form $2^p (2^n - 1) / (2^m - 1)$. Lemma: this is integral if and only if $m | n$. Conclusion: a number is representable if and only if the positions of the 1's digits in its binary representation, form an arithmetic progression. Famous exercise: what is the GCD, $(2^m - 1, 2^n - 1)$?

fleeting_guest wrote: Rushil wrote: least positive integer that cannot be represented in the form $\frac{2^{a} - 2^{b}}{2^{c} - 2^{d}}$ A representable number is of the form $2^p (2^n - 1) / (2^m - 1)$. Lemma: this is integral if and only if $m | n$. Conclusion: a number is representable if and only if the positions of the 1's digits in its binary representation, form an arithmetic progression. Famous exercise: what is the GCD, $(2^m - 1, 2^n - 1)$? If $m,n$ are positive integers and $a>1$ is also an integer, then $gcd(a^m-1,a^n-1)=a^{gcd(m,n)-1}$

In fact, if $a$ and $b$ are coprime, then $\left(a^m - b^m,a^n - b^n\right) = a^{(m,n)} - b^{(m,n)}$

And Prove that it?

Arne, yours is correct To prove this Lemma: $\left(a^m - b^m,a^n - b^n\right) = a^{(m,n)} - b^{(m,n)}$ assume $gcd$ of $(m,n)$ is $k$ $\left(a^m - b^m,a^n - b^n\right) = (a-b) * gcd(\sum a^ib^{m-i-1}, \sum a^ib^{n-i-1}) = (a-b)*sum a^ib^{m-i-1}, \sum a^ib^{n-i-1})$ $a^{(m,n)} - b^{(m,n)} = a^k-b^k = (a-b)(\sum a^ib^{k-i-1})$ $\text{ The value of } \sum a^ib^{m-i-1} \text{ is always a prime number when } m \text{ is odd, and so does n}$ If one of $m$ and $n$ is odd, then $k$ is odd, the lemma is obviously correct If both of them are even, then assume $m=2^{\alpha}m', n = 2^{\beta}n'$where $m', n'$ are all odd, $\alpha,\beta \in N$ then we can prove that the lemma is still true.

You can look here too: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=pgcd&t=36731

Kalimdor wrote: $\text{ The value of } \sum a^ib^{m-i-1} \text{ is always a prime number when } m \text{ is odd, and so does n}$ If one of $m$ and $n$ is odd, then $k$ is odd, the lemma is obviously correct If both of them are even, then assume $m=2^{\alpha}m', n = 2^{\beta}n'$where $m', n'$ are all odd, $\alpha,\beta \in N$ then we can prove that the lemma is still true. What? I don't think this works.

Just download the PDF file and see it !

it was easier to proof c-d|a-b but writing them as binaries was a great idea!!

It's an easy problem. First we say 11=2^a-2^b/2^c-2^d 2^b(2^a-b -1)=11.2^d.(2^c-d -1)==>b=d 11|2^a-n -1 10|a-b==>4|10 So 10 can't divides a-b