random problem i just thought about one day

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ceilingfan404

1139 posts

ceilingfan404

#1 h Apr 20, 2025, 7:54 PM • 4 Y

Y by aidan0626, e_is_2.71828, Exponent11, PikaPika999

i don't even know if this is solvable
Prove that there are finite/infinite powers of 2 where all the digits are also powers of 2. (For example, $4$ and $128$ are numbers that work, but $64$ and $1024$ don't work.)

This post has been edited 1 time. Last edited by ceilingfan404, Apr 20, 2025, 7:55 PM

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huajun78

71 posts

huajun78

#2 h Apr 21, 2025, 12:48 AM • 1 Y

Y by PikaPika999

well as the number gets bigger, there are more digits, so it's less likely that ALL the digits will be a power of 2 (1, 2, 4, 8).

for the first 20 powers of 2 after $2^{10}$ ( $2^{11}$ to $2^{30}$ ), none of them satisfy the condition (I tested all of them), so it's very unlikely that numbers with even more digits will.

I don't know how to prove this but that fact suggests that there are only a finite number.

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vincentwant

1426 posts

vincentwant

#3 h Apr 21, 2025, 1:32 AM • 1 Y

Y by PikaPika999

if the number is greater than 512 then the last four digits must be 2112, 4112, 8112, 2224, 4224, 8224, 1424, 1824, 2144, 4144, 8144, 1184, 2128, 4128, 8128, 1248, 2448, 4448, 8448, 2848, 4848, 8848, 2288, 4288, 8288, 1488, 1888

dont think this helps

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Yummo

299 posts

Yummo

#4 h Apr 21, 2025, 1:36 AM • 1 Y

Y by PikaPika999

@above, what about 1024?

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vincentwant

1426 posts

vincentwant

#5 h Apr 21, 2025, 1:37 AM • 1 Y

Y by PikaPika999

Yummo wrote:

@above, what about 1024?

0 is not a power of 2

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e_is_2.71828

222 posts

e_is_2.71828

#6 h Apr 21, 2025, 1:47 AM • 1 Y

Y by PikaPika999

ceilingfan404 wrote:

i don't even know if this is solvable
Prove that there are finite/infinite powers of 2 where all the digits are also powers of 2. (For example, $4$ and $128$ are numbers that work, but $64$ and $1024$ don't work.)

I won't look into it completely, but we can start somewhere. We'll see if it is possible to "generate" a formula for these numbers. So let $n$ be a $k$ -digit number such that $n=a_ka_{k-1}...a_2a_1a_0$ . Then $n=10^ka_k+10^{k-1}a_k-1...+10a_1+a_0$ , and note for all $i$ $a_i=2^b$ , for some $b$ . So, $n=10^k \cdot 2^{b_k}+10^{k-1}\cdot 2^{b_{k-1}}+...+10\cdot 2^{b_1}+2^{b_0}$ . From there we need also $n=2^c$ for some $c$ , and presumably we can take the largest $b_i$ , factor it out, and we need the remaining sum to also be a power of $2$ . Someone can try working it out from here, I think I started it off well enough.

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wangzrpi

159 posts

wangzrpi

#7 h Apr 24, 2025, 5:06 PM

Y by

See
https://math.stackexchange.com/questions/2238383/how-many-powers-of-2-have-only-0-or-powers-of-2-as-digits

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e_is_2.71828

222 posts

e_is_2.71828

#8 h Apr 24, 2025, 5:13 PM

Y by

Definitely not middle school math

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e_is_2.71828

222 posts

e_is_2.71828

#10 h Apr 24, 2025, 6:00 PM • 1 Y

Y by PikaPika999

K1mchi_ wrote:

e_is_2.71828 wrote:

Definitely not middle school math

its fine

doesn’t need to be msm curriculum
just for msm

if u can’t do it skill issue

"This problem is unlikely to have a simple proof, because the following holds:

Theorem. For any k, there exists a power of 2 whose first k digits and last k digits are all either 1 or 2.
Proof. We begin with looking at the last digits, taking 2nmod10k. For sufficiently large n, 2n≡0(mod2k). Since 2 is a primitive root modulo 5 and modulo 52, it is a primitive root modulo 5k for any k (Wikipedia), so we can have 2n≡b(mod10k) for any b such that b≡0(mod2k).

This is possible to accomplish with only 1 and 2 as digits. We start with b1=2 for k=1, and extend bk−1≡0(mod2k−1) to bk≡0(mod2k) by the rule:

If bk−1≡0(mod2k), take bk=2⋅10k−1+bk−1.
If bk−1≡2k−1(mod2k), take bk=10k−1+bk−1.
(This works because 10k−1≡2k−1(mod2)k.)

There is a unique sequence of digits ending …211111212122112 that we obtain in this way; reversed, it is A023396 in the OEIS.

To make sure that 2n ends in bk, there will be some condition along the lines of
n≡c(modϕ(5k))
or n=c+n′ϕ(5k) for some n′. From there, getting the first k digits to be 1 or 2 is easy along the lines of a recently popular question. We might as well aim for the sequence 111…111k, because we can. To do this, we want
log101.11…1<{(c+n′⋅ϕ(5k))log102}<log101.11…2
where {x} denotes the fractional part of x. This translates into a condition of the form
{n⋅log102ϕ(5k)}∈Ik
for some interval Ik, which we know is possible because α=log102ϕ(5k) is irrational, and therefore the sequence {α},{2α},{3α},… is dense in [0,1].

This concludes the proof.

Instead of the digits {1,2} we could have used the digits {1,4} or {1,8} and given a similar proof; if we multiply the solution to one of these by 2 or 4, we get a power of 2 whose first and last digits come from the set {2,4} or {2,8} or {4,8}. (We can't do this with just the set {0,1} or {0,2} or {0,4} or {0,8}, because eventually we can rule these out by a modular condition.)

It's of course still almost certain that there's no large power of 2 entirely made from the digits {0,1,2,4,8}, but you'd have to say something about the "middle digits" of such a power, which is much harder."

From the stack exchange.

This post has been edited 1 time. Last edited by e_is_2.71828, Apr 24, 2025, 6:01 PM
Reason: Added

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Andrew2019

2312 posts

Andrew2019

#12 h Apr 25, 2025, 4:42 PM • 2 Y

Y by e_is_2.71828, Demetri

K1mchi_ wrote:

e_is_2.71828 wrote:

Definitely not middle school math

its fine

doesn’t need to be msm curriculum
just for msm

if u can’t do it skill issue

it would be crazy if someone who has only done the amc 8 and sold on it says others have a skill issue

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maromex

186 posts

maromex

#13 h Apr 25, 2025, 5:08 PM

Y by

There is a related question: Does the base- $3$ expression of $2^n$ always have a digit equal to $2$ for sufficiently large $n$ ? If I recall correctly, this problem is unsolved.

The problem discussed in this topic seems similar to this question, and I don't see why it would be solvable with currently known techniques.

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e_is_2.71828

222 posts

e_is_2.71828

#14 h Apr 25, 2025, 5:15 PM • 1 Y

Y by mithu542

I wouldn't listen to someone who can't even spell figure ...

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maxamc

576 posts

maxamc

#15 h Apr 25, 2025, 5:22 PM

Y by

e_is_2.71828 wrote:

I wouldn't listen to someone who can't even spell figure ...

K1mchi_ is always right 100000 aura.

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K1mchi_

85 posts

K1mchi_

#16 h Apr 25, 2025, 5:39 PM

Y by

Andrew2019 wrote:

K1mchi_ wrote:

e_is_2.71828 wrote:

Definitely not middle school math

its fine

doesn’t need to be msm curriculum
just for msm

if u can’t do it skill issue

it would be crazy if someone who has only done the amc 8 and sold on it says others have a skill issue

slander

i just dont do competitive math

hate me if u like

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MathPerson12321

3799 posts

MathPerson12321

#17 h Apr 25, 2025, 5:52 PM • 2 Y

Y by e_is_2.71828, mithu542

#11
why dont u?

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K1mchi_

85 posts

K1mchi_

#18 h Apr 25, 2025, 7:07 PM

Y by

MathPerson12321 wrote:

#11
why dont u?

just quote me

i have better things to do with my time than math rn

i’ll do u the service of enlightenment if i ever find time

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Aaronjudgeisgoat

899 posts

Aaronjudgeisgoat

#19 h Apr 25, 2025, 7:12 PM

Y by

K1mchi_ wrote:

MathPerson12321 wrote:

#11
why dont u?

just quote me

i have better things to do with my time than math rn

i’ll do u the service of enlightenment if i ever find time

you only have 105 posts, but i feel like ive seen you everywhere

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MathPerson12321

3799 posts

MathPerson12321

#22 h Apr 25, 2025, 8:43 PM

Y by

@bove stop trying to say ur better
do i see mop quals trying to bring me down? no

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RollingPanda4616

258 posts

RollingPanda4616

#24 h Apr 25, 2025, 10:56 PM • 1 Y

Y by PikaPika999

K3mchi_ wrote:

MathPerson12321 wrote:

@bove stop trying to say ur better
do i see mop quals trying to bring me down? no

so? im not trying to bring u down u still bring urself down bc ur very sensitive

dont we celebrate intelligence in our society?

alt alert

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RollingPanda4616

258 posts

RollingPanda4616

#26 h Apr 26, 2025, 5:10 PM • 2 Y

Y by PikaPika999, e_is_2.71828

hey

yes
$~~~~~~$

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valisaxieamc

442 posts

valisaxieamc

#27 h Apr 26, 2025, 5:19 PM • 3 Y

Y by RollingPanda4616, PikaPika999, e_is_2.71828

Bro imagine making alts cause you fear that aops is going to ban you

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RollingPanda4616

258 posts

RollingPanda4616

#28 h Apr 26, 2025, 5:27 PM • 1 Y

Y by PikaPika999

valisaxieamc wrote:

Bro imagine making alts cause you fear that aops is going to ban you

anyway let's get this thread back on track

I think you might need to break up the digits and use the prime factorization. (like a 3 digit number $abc$ would be broken down into $a \cdot 2^2 5^2 + b \cdot 2^1 5^1 + c$ and since a, b,c are powers of 2, you could just look at the 5s?) idk how to continue though.

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maromex

186 posts

maromex

#29 h Apr 26, 2025, 7:15 PM

Y by

I'll say this again:

maromex wrote:

There is a related question: Does the base- $3$ expression of $2^n$ always have a digit equal to $2$ for sufficiently large $n$ ? If I recall correctly, this problem is unsolved.

The problem discussed in this topic seems similar to this question, and I don't see why it would be solvable with currently known techniques.

Unless a problem about digits has good reason to be solvable with currently known techniques, it's probably not solvable, even if the answer seems obviously true/false at first.

Here's another unsolved problem related to the topic of this thread: For $n > 86$ , does $2^n$ always have a $0$ in base $10$ ?

This post has been edited 1 time. Last edited by maromex, Apr 26, 2025, 7:16 PM

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PikaPika999

1781 posts

PikaPika999

#30 h Apr 27, 2025, 11:08 PM

Y by

K1mchi_ wrote:

e_is_2.71828 wrote:

Definitely not middle school math

its fine

doesn’t need to be msm curriculum
just for msm

if u can’t do it skill issue

but if the forum is literally called msm, then shouldn't it be msm? plus, if it is harder than msm, there are high school math and college math and high school olympiads, and it could've been placed there?

k1mchi_

not nice

valisaxieamc wrote:

Bro imagine making alts cause you fear that aops is going to ban you

lol imo aops should use ip bans

This post has been edited 1 time. Last edited by PikaPika999, Apr 27, 2025, 11:09 PM

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PikaPika999

1781 posts

PikaPika999

#31 h Apr 27, 2025, 11:14 PM • 2 Y

Y by RollingPanda4616, Pengu14

K3mchi_ wrote:

MathPerson12321 wrote:

@bove stop trying to say ur better
do i see mop quals trying to bring me down? no

so? im not trying to bring u down u still bring urself down bc ur very sensitive

dont we celebrate intelligence in our society?

1. True intelligence shines through clarity and simplicity, not overcomplication.
2. Intelligence isn’t just about flaunting knowledge—it’s also about understanding, humility, and connection.
3. True intelligence lies not in power over others, but in empowering those around us.
4. Creativity/intelligence isn’t just about thinking outside the box—it’s about reshaping the box entirely.
5. Leadership isn’t a title—it’s the trust you earn and the influence you wield wisely.
6. Intelligence is not in the answers we give, but in the questions we dare to ask.
7. Intelligence grows when we challenge our own assumptions, not just those of others.
8. The hallmark of intelligence is recognizing that there’s always more to learn.
9. Intelligence flourishes in collaboration, not isolation.

This post has been edited 2 times. Last edited by PikaPika999, Apr 27, 2025, 11:15 PM

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valisaxieamc

442 posts

valisaxieamc

#33 h Apr 29, 2025, 5:01 AM • 1 Y

Y by PikaPika999

I completely agree with PikaPika but like RollingPanda said, we probably should get back on topic. I mean the kimchi dude is finally leaving us alone and hopefully getting a life so I'll take it as a win

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fake123

93 posts

fake123

#34 h Apr 29, 2025, 6:15 AM

Y by

bro why are you guys raging over some random kid why can't you just ignore him

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PikaPika999

1781 posts

PikaPika999

#35 h Apr 29, 2025, 11:26 AM

Y by

fake123 wrote:

bro why are you guys raging over some random kid why can't you just ignore him

we're not raging over "some random kid" who can be ignored (sorry if this sounds harsher than it is)

they start flamewars on multiple different threads. This is how my 1000th post thread got locked :furious

also, they created multiple different alts, which is explicitly said to be against the rules (probably because of getting postbanned from this sheriff

sry if this sounds harsher than i meant to be

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