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# Is \$2^{16} = 65536\$ the only power of \$2\$ that has no digit which is a power of \$2\$ in base-\$10\$?

13 I was watching this video on YouTube where it is told (at 6:26) that \$2^{16} = 65536\$ has no powers of \$2\$ in it when represented in base-\$10\$. Then he – I think as a joke – says “Go on, find another power of \$2\$ that doesn’t have a power of \$2\$ digit within it. I dare you!”

So I did. 🙂 I wrote this little Python program to check for this kind of numbers:

You are watching: 65536*2

toThePower = 0 possiblyNoPower = True while True: number = str(2**toThePower) for digit in number: if int(digit) in [1,2,4,8]: possiblyNoPower = False print(‘Not ‘ + number) break if possiblyNoPower: print(number + ‘ has no digit that is a power of 2.’) toThePower += 1 possiblyNoPower = True

Sidenote: I could use the programming language Julia instead of Python, which may be much quicker, but I already checked for really big numbers and such a program (and brute-force in general) will never proof that there are no other powers of \$2\$ having this property. It might disprove it, but I think the chance is really really small.

See more: Jigsaw Block | Pink Army

I checked all the way to \$2^{23826}\$, which is a 7173 digit number, but no luck. Since the numbers are getting more and more digits with bigger powers of \$2\$, the chance of a number having no digit that is a power of \$2\$ becomes smaller and smaller.

I made a plot of \$frac{text{number of digits being a power of 2}}{text{total number of digits}}\$ versus the \$n\$th power of \$2\$ on a logarithmic scale. This graph is wrong! See the edit.

As I predicted, the graph drops really fast to almost \$0\$. I think that \$frac{text{number of digits being a power of 2}}{text{total number of digits}}rightarrow 0\$ as \$n rightarrow infty\$ and thus \$text{P}(ntext{th power of 2 having no digits of 2 in it}) rightarrow 0\$, but this is just my intuition.

So my question: Is \$2^{16} = 65536\$ the only power of \$2\$ that has no digit in it that is a power of \$2\$ (so no digit \$in {1,2,4,8}\$) when represented in base-\$10\$? Is there a proof, a counterexample, or is it an open question? I’m also curious about powers of \$2\$ having no digits that are a power of \$2\$ in other bases than \$10\$.

Edit: As @Aweygan noted, the graph above is wrong. I accidentally divided by the number itself, instead of the amount of digits the number has. Below a good version, on a linear scale. From this graph it appears that \$frac{text{number of digits being a power of 2}}{text{total number of digits}} rightarrow 0.4 \$ as \$n rightarrow infty\$. This seems to make sense, since \$text{P}(text{digit} in {1,2,4,8}) = 4/10 = 0.4\$ and since the number of digits becomes larger and larger, the law of large numbers becomes “visible”.

About the possible duplicate: that question was posted 4.5 years ago. The status of the problem (proven, disproven, open question) might well be changed in the maintime. 🙂

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