[eqn] α = 2^n where α is also divisible by 3 [/eqn]
Is this possible? Explain and discuss as intuitively as possible.
[eqn] α=2n [/eqn]
where α is also divisible by 3
>>8552786
Fucking hell [eqn] α=2^n [/eqn]
where where α is also divisible by 3
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no because 2*2*2*... can be viewed as a prime factorisation of a and it will never have any 3 in there
https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic#Canonical_representation_of_a_positive_integer
>>8552785
use fibonacci you fucktard
>>8552785
n=ln3/ln2
>>8552791
No, because 2^n has unique prime factorization 2^n, therefore alpha can't be a multiple of 3.
>>8552785
If n is an integer, then no.
>>8552819
you could have an irrational answer but it wouldn't be an integer.