$2^k$ not divisible by 5 In Knuth's Art of Computer Programming, Volume 1 it is stated without proof that $2^k$ is not divisible by 5 (answer to question 10, section 1.2.2).  I can construct a short proof, but what intuition am I missing that makes this so obvious?
 A: The easiest way is to spot that the only prime factor of $2^k$ is 2, while it needs to have $5$ as prime factor in order to be divisible by 5.
Also you can check that all powers of 2 end in $2,4,8$ or $6$, meaning it never ends in $0$ or $5$, hence the proof.
A: Since $5$ is prime, it would have to appear in a prime factorization of any number it divides... but you already know a prime factorization of $2^k$ without $5$!
A: It is an immediate consequence of the uniqueness of the prime factorization $\,2^k = 2\cdot 2\,\cdots\, 2$
More simply $\ \color{#C00}5\mid \color{#C00}{2n}\, \Rightarrow\, \color{#C00}5\mid (\color{#c00}5n - 2\!\cdot\! \color{#c00}{2n}) = n.\,$ By induction $\, 5\mid 2^km \,\Rightarrow\, 5\mid m.$
That is a special case of Euclid's Lemma $\, a\mid bc\,\Rightarrow\, a\mid (a,b)c,\, $  the case $\,(a,b) = 1.$ Euclid's Lemma follows immediately from Bezout's identity for the gcd $\,(a,b) = ja+kb\,$ for some $\,j,k\in\Bbb Z.\,$ By the special case of $\,a\,$ prime in Euclid's Lemma, $\,p\mid bc\,\Rightarrow\, p\mid b\,$ or $\,p\mid b,\,$ since $\,p\nmid b\,$ $\Rightarrow$ $\,(p,b)=1,\,$ so $\,p\mid bc\overset{\rm Euclid}\Rightarrow p\mid(p,b)c = c.\,$ This Prime Divisor Property yields  uniqueness of prime factorizations, since it permits one to match-up and cancel primes in any two prime factorizations of an integer.
A: If $p$ is prime then $p\mid k\times m$ implies that $p\mid k$ or $p\mid m$. Applying (and repeating) on $5\mid 2^n$ leads to $5\mid 2$ wich is not true.
