Tribonacci Exponents!

Number Theory Level 2

The following powers of 2 consist of all even digits:

$\begin{array}{lclcr} & & 2^{\color{#D61F06}1} &= &2\\ & & 2^{\color{#D61F06}2} &= & 4\\ 2^{\color{#D61F06}3} &= &2^{\color{#D61F06}{1+2}} &= & 8\\ 2^{\color{#D61F06}6} &= &2^{\color{#D61F06}{1 + 2 + 3}} &= & 64\\ 2^{\color{#D61F06}11} &= &2^{\color{#D61F06}{2 + 3 + 6}} &= & 2048 \end{array}$

Does $2^{\color{#D61F06}3+6+11}$ contain all even digits?

Generalization proofs are more than welcome here.

Yes No

1 solution

X X
Aug 2, 2018

[This isn't a generalization proof.]

$2^{3+6+11}=2^{20}=1048576$

I'm not sure what $2^n$ will contain only even digits,but we can look for $2^{n-1}$ which contains digits $0,1,2,3,4$ only. ( I guess there are no more after 2048)

View OEIS

There has been a discussion about $2048$ known for having the even digits. The fact that there is no obvious generalization relies very deep on the multiplication. However, this isn't enough to cover the proof. Even some articles explore some parts of powers of $2$ via high-level engineering methods.

Here is the discussion from Stack Exchange I am referring to.

Michael Huang - 2 years, 10 months ago

Tribonacci Exponents!

1 solution

0 pending reports