1. ## proof required....

can somebody plz find a value or show me how to find a value for x such tht,

$2^x/3 = n$

where x and n are integers

or otherwise prove tht for no integeral value of x can $2^x$ be an integeral multiple of 3.

2. An induction proof should work well with this.

An observation to make is that powers of 2 alternate between congruencies depending if n is odd or even.

$2^{n}\equiv{1}(mod \;\ 3)$, if n is even

$2^{n}\equiv{-1}(mod \;\ 3)$, if n is odd.

That is, $2^{n}-1$ divides 3 if n is even and $2^{n}+1$ divides 3 if n is odd.

3. Originally Posted by maliksaim
can somebody plz find a value or show me how to find a value for x such tht,

$2^x/3 = n$

where x and n are integers

or otherwise prove tht for no integeral value of x can $2^x$ be an integeral multiple of 3.
Not sure what class you are in exactly, but
$\frac{2^x}{3} = n$

$2^x = 3n$

Now, if x and n are integers, then 2^x and 3n are integers.

So look at the prime factorization of both sides. Clearly the prime factorization of the RHS contains a 3, but the prime factorization of the LHS contains only 2s. Thus the two sides cannot be equal for x and n integers.

-Dan

4. Originally Posted by maliksaim
can somebody plz find a value or show me how to find a value for x such tht,

$2^x/3 = n$

.
It is not possible we require that $3 | 2^x$ which is simply impossible.

5. Originally Posted by maliksaim
[snip]
or otherwise prove tht for no integeral value of x can $2^x$ be an integeral multiple of 3.
Originally Posted by ThePerfectHacker
[snip]
we require that $3 | 2^x$ which is simply impossible.
Gets my vote for proof of the year!!

6. Originally Posted by mr fantastic
Gets my vote for proof of the year!!
The question was stupid anyway.

7. yea the question seemed stupid, but it was really required for something very interesting, the MIU formal system from "godel, eshcer, bach". i thought i'd give it a try. i guess i just wasnt meant to take the glory of proving it