All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
521st Digit Determination (Posted on 2023-07-28) Difficulty: 3 of 5
Determine the 521st digit to the right of the decimal point in the base ten representation of:
       (1+√2)2022

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution computer solution and an observation | Comment 1 of 2
The digit is 9.

>> digits 4000
>> (1+sqrt(vpa(2)))^2022
ans =
9385031900421997499757131428491061514215164000646844276126035536184226698323836996059437908127640457
7956284443582890211201374006802872246511586219630281769959869305566691011377516678008285481757698513
3628163002589205672656801331777149834581515352832012325992401047082314593220792999314472113925777452
7272330513732920927023283924308688056246650479193807532300881937910688213376157047347667044032722544
9276511065038655540170977801608689308393295333424979118410410531622297175911701560783394613400400049
8026307803154107020338784709272977506562483251491520944518124934623249229159409621742226468150354305
7762679465348434025228014552152567459273269085203066231016626377266314008191422732267623040339675694
91150413287891344696018460312441258110126250727310293603443121951886073797.9999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999998934473520590766400083916087927364285152177744625304
8705400306190952722781177167962344213748388552130224524640505207466841350467228964543266517619174527
2319635820147746802454803121618411007085067217063844889187901801397331783208665861711760818945402398
7453447844716712678539664611066849050860631701163375482353838180338330057960979315570871431112418801
1432846842258505280906743122515453112

The digits that are not 9 don't begin until the 776th digit to the right of the decimal, so the 521st position is one of the long string of 9's.

If the power had been 2023, the answer would have been zero:

It's interesting that powers of 1+sqrt(2) and 1+sqrt(3) both go into an alternation between just above and integer and just below an integer (many 0's alternating with many 9's). This doesn't seem to happen with just any irrational numbers, such as pi or 1+sqrt(7), or sqrt(7) itself. I don't know what's special about these two (and I don't know how many more) that do jump about integral values).

  Posted by Charlie on 2023-07-28 09:31:58
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (8)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information