What is the lowest positive integer multiple of $7$ that is also a power of $2$ (if one exists)? 
What is the lowest positive multiple of $7$ that is also a power of $2$ (if one exists)?

Not a homework question, I am not in school, I am just wondering what the answer is.
 A: There is none. The only divisors of a number of the type $2^n$ are of the form $2^m$. 
One way of seeing this is that powers of 2 will always leave remainders of either 1,2,or 4 when divided by 7:
$2^1$ leaves a remainder of 2
$2^2$ leaves a remainder of 4
$2^3$ leaves a remainder of 1
You can then see that any other power of 2 will leave a remainder of either 1,2, or 4, when dividing by 7,so that , when using the division algorithm a=pq+r with a=2^n and p=7, r will never be 0, which is what you want for 7 (and so a multiple of 7) to divide $2^n$ .
A: We prove a result that answers the question and shows a little more. Let $7k$ be a positive multiple of $7$. Then $\log_2(7k)$ is irrational. 
Suppose to the contrary that $\log_2(7k)=\frac{a}{b}$, where $a$ and $b$ are integers. Then 
$$7k=2^{a/b} \qquad\text{and therefore}\qquad (7k)^b=2^a.$$
This is impossible, since $7$ divides $(7k)^b$ but $7$ does not divide $2^a$. The reason is that if a prime (in this case $7$) divides a product, then it divides one of the terms.  However, $7$ does not divide $2$.
In particular, the proof shows that $\log_2(7k)$ is not an integer, that is, that $7k$ is not an integer power of $2$.
There is nothing particularly special here about $7$ and $2$. The interesting thing is that the argument shows, in a very simple way, that for example $\log_2(7)$ is irrational. Results of this type are among the simplest to prove irrationality results.
A: No power of 2 is a multiple of 7; by the uniqueness of prime factorization, a number $n$ cannot have two prime factorizations 
$$n=2^a$$
and
$$n=7^{b}p_1^{c_1}p_2^{c_2}\cdots p_r^{a_r}$$
where $b\geq1$, because the factorizations must be different (the power of 7 occuring in the first is $7^0=1$, while the power of 7 occuring in the second is $7^b$ where $b\geq1$). Thus, a number cannot be both a power of 2 and divisible by 7.
