Is it possible to prove this?

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Thales12345

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I was wondering if it is possible to prove that there is/isn't a natural number such that the number raised to the power of 2021 starts with 2021
 
Thales12345 said:
I was wondering if it is possible to prove that there is/isn't a natural number such that the number raised to the power of 2021 starts with 2021
Click to expand...
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
I don’t even know of this is possible and that is my question. If so, how can we prove this? I was thinking about logarithm.
 
There are 23 solutions less than 100,000. To find them, I wrote a computer program that looped through all numbers from 2 to 99,999. Within the loop, a simple test involving logarithms is used to determine if "x" is a solution or not. Can you work out the test? Here's a pretty big clue...

log10(2021) = 3.30556631351530399082
log10(20210) = 4.30556631351530399083
log10(202100) = 5.30556631351530399083		log10(2022) = 3.30578115125498225870
log10(20220) = 4.30578115125498225871
log10(202200) = 5.30578115125498225871
 
Cubist,
The way I fiddled with this problem did not involve any base.
The OP wants an x value in the natural numbers such that x^2021 = 2021x...x.
 
Jomo said:
Cubist,
The way I fiddled with this problem did not involve any base.
Click to expand...
Hopefully you've found a better way to find a solution (solutions). I didn't particularly like having to test lots of values in my program to see which ones fit - it doesn't seem very elegant or mathematical.

Jomo said:
The OP wants an x value in the natural numbers such that x^2021 = 2021x...x.
Click to expand...
I think I've understood the problem. x*x*x*...*x*x with 2020 multiplications
 
I extended the spreadsheet and found 251 such integers in the first million, the largest being 996935.
I also found that every 4 digit number appears as the 'start' of the decimal form of [MATH]n^{2021}[/MATH]
Subhotosh Khan said:
How did you tame that beast?
Click to expand...
I calculated 2021*log(n). If I call this x, I then calculated 10^(x-int(x)), multiplied this by 1000 and took the integer part.
 
Let x = the natural number we want to find.

x2021 = 2021x...x then x = 2021x...x1/20021
It seems that x is bnd by 1 and 2 never reaching either.
 
This is what I was hinting at in post#5...

log10(2021) ≤ 2021*log10( x ) - n < log10(2022)

where n,x any integer > 0. This led to the code below, which shows single values, and ranges, for x (looping through different n values).

--

This is the first of infinite ranges of consecutive numbers that provide a solution...
4334756 to 4334757

And here's two more ranges...
9132020 to 9132021
...
1000348201349446861367010426331 to 1000348446205699389933132707976
...

From these, you can get...

9132021 ^ 2021 = 2021659348...

1000348202120212021202120212021 ^ 2021 = 2021003147...

...but I don't think any x could begin with the digits 2021

Code: 
n=0;
loop(forever) {
  l = floor(pow(10, (log10(2021) + n)/2021) + 1);
  h = floor(pow(10, (log10(2022) + n)/2021));
 
  if (h == l) {
    print "x=", l;
  } else if (h >= l) {
    print l, " <= x <= ", h;
  }
 
  n += 1;
}
[code]
[/SPOILER]
 
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Jomo said:
x2021 = 2021x...x then x = 2021x...x1/20021
Click to expand...

I only just got the 2021x...x notation! The "x...x" represents unknown digits that complete the number! Doh, maybe I shouldn't admit to being slow :oops::LOL:
 
@Cubist
Does your code not first identify n=446 as a number satisfying the requirement (as my spreadsheet did)?
And how do you know there are infinitely many ranges containing such integer solutions?
 
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I have it now. Yes there is an infinite number of solutions.
We simply want there to be an integer in between [MATH]2.021^{1/2021}(10^{1/2021})^k[/MATH] and [MATH]2.022^{1/2021}(10^{1/2021})^k[/MATH].
Now eventually the difference between these two numbers becomes greater than 1, therefore there must be an integer between them.
Subtracting the two numbers we get: [math](2.022^{1/2021}-2.021^{1/2021})(10^{1/2021})^k[/math] which is >1 when [math]k>\dfrac{ \log \dfrac{1}{2.022^{1/2021}-2.021^{1/2021}} } {\log(10^{1/2021})}[/math] which is approx 13361.

This can generalise for other numbers.
E.g. to find one of the infinite number of solutions to [math]n^{161}=161....[/math]take any value of k≥711 and let [math]n=\lfloor 1.62^{1/161} 10^{k/161}\rfloor[/math]
 
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In the last example, perhaps better to take [math]n=\lceil 1.61^{1/161} 10^{k/161}\rceil[/math] just in case the other is an integer!
 
lex said:
@Cubist
Does your code not first identify n=446 as a number satisfying the requirement (as my spreadsheet did)?
Click to expand...
Yes it does output the first number, 446. Sorry, I probably confused things because my method changed between writing post#5 and post#12, and I didn't explicitly say so. My initial method (in post#5) functioned in the same way as your spreadsheet. EDIT: But I see that you have it now!

Anyway, @Thales12345 , you have started two very interesting threads! You might like the following, although they're very ugly if you look at ALL the digits of the various results...

9999 ^ 99 = 99...
156156 ^ 156 = 156...
216216 ^ 216 = 216...
414414 ^ 414 = 414...
999999 ^ 999 = 999...
17851785 ^ 1785 = 1785...
99999999 ^ 9999 = 9999...
5102451024 ^ 51024 = 51024...
 
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lex said:
We simply want there to be an integer in between [MATH]2.021^{1/2021}(10^{1/2021})^k[/MATH] and [MATH]2.022^{1/2021}(10^{1/2021})^k[/MATH].
Click to expand...
So this difference can be more than 1. Wolfram Alpha did not seem to show that. I should have investigated further. (Subhotosh) I am guilty of sloppy thinking! I never should have assumed anything just from partial information from a calculator.
 
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