$2^{61}$ as a multiple of 3... is that possible?

Question

I had a recent conversation with a friend of mine who said that $2^{61}$ was a multiple of 3, but I wanted to disprove this argument by claiming that all values of $2^n$ were not a multiple of 3 at all, and that it was impossible for such a claim to exist.

Is $2^{61}$ a multiple of 3 and why?

Thanks.

Answer 1

The reason is

$$ 2 \ \mathrm{mod}\ 3 = -1 \implies 2^n \ \mathrm{mod}\ 3= (-1)^n=-1 \text{ or } 1 = 2 \text{ or } 1 $$

Answer 2

What lulu commented: Every integer can be uniquely separated into a multiplication of primes called a "factorization" (which is just the number itself, if it is a prime). The integer is only divisible by the primes that are part of the factorization. $2^n$ means that the factorization is $2\cdot2\cdot2\cdot2\cdot2\cdot...$ . Hence the number is divisible by no other prime than $2$.

Answer 3

No $2^n$ is not divisible by 3 as it does not contains any factor of $3$

So $2^{61}$ is not divisible by 3

Answer 4

My own answer

$2^{61} = 2 * 2* 2* 2... (61 \ times)$

There is no factor of 3, so this is impossible.