Interesting algebra involving the square root of a summation

Jan 2013
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0
Let \(\displaystyle N=1+10+10^2+\cdots+10^{4023}\). Find the 2013-th digit after the decimal comma of \(\displaystyle \sqrt{N}\).
 
Jul 2010
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St. Augustine, FL., U.S.A.'s oldest city
Re: Interesting algebra involving the square root of a summa

Where did you find this problem?
 

greg1313

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Oct 2008
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London, Ontario, Canada - The Forest City
Re: Interesting algebra involving the square root of a summa

1
 

mathbalarka

Math Team
Mar 2012
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Re: Interesting algebra involving the square root of a summa

roberthun said:
Let \(\displaystyle N=1+10+10^2+\cdots+10^{4023}\). Find the 2013-th digit after the decimal comma of \(\displaystyle \sqrt{N}\).
Since the series is a geometric series, N = (1 - 10^4024)/(1 - 10). 10^4024 = 10000....000 implies, 1 - 10^4024 = - (10^4024 - 1) = - 999999.....999. Hence we get N = -999.....999/(-9) = 999....999/9 = 111.....111.

I have no idea what to do about sqrt(N).

greg1313 said:
This is almost surely false. Maybe you meant that the 2013-th digit of N. but OP asked for sqrt(N).
 

mathbalarka

Math Team
Mar 2012
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India, West Bengal
Re: Interesting algebra involving the square root of a summa

Aha! I will do arithmetic on the numbers of the form 11.....111 (agentredlum discovered such numbers in a very old topic : viewtopic.php?f=40&t=18441).

the agenterdlum number sqrt(111.....11) can be mapped in the reals by transforming sqrt(11...1) to sqrt(1.111.....) where the place of the decimal doesn't matter. Now, sqrt(1.111....) = sqrt(11(0)/99) where 11(0) means there can be uncountably/countably many zeros after 11 (i.e. 110, 1100,...). we will take 11(0) = 11 for our computational matters. sqrt(11/99) = sqrt(11)/3sqrt(11) = 1/3 = 0.333333.... which is in turn represents the agentredlum number 3333.....333. Hence, no matter where the decimal is situated, the next number after the decimal is 3, and so as the next and next and e.t.c. Hence, the 2013-th digit (irrelevant) after the decimal (irrelevant,too) is always 3.

NOTE : @agentredlum, seems like we find a practical application of your idea :D I must admit, you're just brilliant!!

Balarka

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greg1313

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Oct 2008
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London, Ontario, Canada - The Forest City
Re: Interesting algebra involving the square root of a summa

mathbalarka said:
This is almost surely false.
:evil: It's not false, it is 1!

roberthun said:
Let \(\displaystyle N=1+10+10^2+\cdots+10^{4023}\). Find the 2013-th digit after the decimal comma of \(\displaystyle \sqrt{N}\).
\(\displaystyle \sqrt{11}\,\approx\,3.31\) (accurate to 2 decimal places).

\(\displaystyle \sqrt{1111}\,\approx\,33.331\) (accurate to 3 decimal places).

\(\displaystyle \sqrt{111111}\,\approx\,333.3331\) (accurate to 4 decimal places).

. . . and so on.
 

mathbalarka

Math Team
Mar 2012
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India, West Bengal
Re: Interesting algebra involving the square root of a summa

greg1313 said:
:evil: It's not false, it is 1!
I agree that the digit 1 is in the decimal expansion but how would you guarantee that '1' is the 2013-th digit of the decimal expansion of sqrt(111....1(4022 times)) ? You can't prove such arguments by creating analogues for N.
 
Mar 2012
654
11
Belgium
Re: Interesting algebra involving the square root of a summa

\(\displaystyle sqrt(1111)=33.33166=33.332\)
and not \(\displaystyle 33.331\)
 

mathbalarka

Math Team
Mar 2012
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India, West Bengal
Re: Interesting algebra involving the square root of a summa

gelatine1 said:
\(\displaystyle sqrt(1111)=33.33166=33.332\)
and not \(\displaystyle 33.331\)
He gave an approximation accurate to 3 decimal places not truncating the last digit.
 
Mar 2012
654
11
Belgium
Re: Interesting algebra involving the square root of a summa

but even though it is 1.
\(\displaystyle sqrt(11(2times))= 3(1time).3(1time)16 =3.316
sqrt(11(4times))= 3(2times).3(2times)16=33.3316
sqrt(11(6times))= 3(3times).3(3times)16=333.33316\)
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\(\displaystyle sqrt(11(4024times))= 3(2012times).3(2012times)16\)
so 2013th digit after comma is 1.