can somebody plz find a value or show me how to find a value for x such tht,
$\displaystyle 2^x/3 = n$
where x and n are integers
or otherwise prove tht for no integeral value of x can $\displaystyle 2^x$ be an integeral multiple of 3.
can somebody plz find a value or show me how to find a value for x such tht,
$\displaystyle 2^x/3 = n$
where x and n are integers
or otherwise prove tht for no integeral value of x can $\displaystyle 2^x$ be an integeral multiple of 3.
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.
$\displaystyle 2^{n}\equiv{1}(mod \;\ 3)$, if n is even
$\displaystyle 2^{n}\equiv{-1}(mod \;\ 3)$, if n is odd.
That is, $\displaystyle 2^{n}-1$ divides 3 if n is even and $\displaystyle 2^{n}+1$ divides 3 if n is odd.
Not sure what class you are in exactly, but
$\displaystyle \frac{2^x}{3} = n$
$\displaystyle 2^x = 3n$
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