# 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.

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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 $2^n$ .

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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.

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In other words, a power of 2 factors as 2 to some power (of course), so there are only 2's present. Thus, it cannot be that it is also a multiple of 7, because this would require a 7 to be present in the factorization. – Austin Mohr Oct 22 '11 at 0:14
@Zev : I had started I new account under "Gary" because my computer crashed, and I lost my login info. I had left a message for Willie a while back, but maybe he did not get it. I am trying to consolidate both accounts into one. Would you please help me do that? – gary Oct 22 '11 at 0:20
@gary: Could you point me to a post from the other account? I can't seem to find it. – Zev Chonoles Oct 22 '11 at 0:25
@gary, Zev: I believe that math.stackexchange.com/users/17850/gary is the intended account. – Asaf Karagila Oct 22 '11 at 0:46
@Asaf: Thanks for the help, I've now merged the accounts. – Zev Chonoles Oct 22 '11 at 0:50

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.

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