Thales12345
New member
- Joined
- Jul 3, 2021
- Messages
- 30
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
Please show us what you have tried and exactly where you are stuck.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
log10(2021) = 3.30556631351530399082 log10(20210) = 4.30556631351530399083 log10(202100) = 5.30556631351530399083 | log10(2022) = 3.30578115125498225870 log10(20220) = 4.30578115125498225871 log10(202200) = 5.30578115125498225871 |
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.Cubist,
The way I fiddled with this problem did not involve any base.
I think I've understood the problem. x*x*x*...*x*x with 2020 multiplicationsThe OP wants an x value in the natural numbers such that x^2021 = 2021x...x.
How did you tame that beast?n=446
(Thanks to Excel spreadsheet)!
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.How did you tame that beast?
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]
x2021 = 2021x...x then x = 2021x...x1/20021
What admit? We knew that all along.......I shouldn't admit to being slow
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!@Cubist
Does your code not first identify n=446 as a number satisfying the requirement (as my spreadsheet did)?
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.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].