Let #N# be the positive integer with #2018# decimal digits, all of them 1: that is #N = 11111cdots111#. What is the thousand digit after the decimal point of #sqrt(N)#?

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Answer 1 · 2017-06-29T18:50:47

#3#

Note that the given integer is #1/9(10^2018-1)#, so it has positive square root very close to #1/3(10^1009)#

Note that:

#(10^1009-10^-1009)^2 = 10^2018-2+10^-2018 < 10^2018-1#

#(10^1009-10^-1010)^2 = 10^2018-2/10+10^-2020 > 10^2018-1#

So:

#10^1009-10^-1009 < sqrt(10^2018-1) < 10^1009-10^-1010#

and:

#1/3(10^1009-10^-1009) < sqrt(1/9(10^2018-1)) < 1/3(10^1009-10^-1010)#

The left hand side of this inequality is:

#overbrace(333... 3)^"1009 times".overbrace(333...3)^"1009 times"#

and the right hand side is:

#overbrace(333... 3)^"1009 times".overbrace(333...3)^"1010 times"#

So we can see that the #1000#th decimal place is #3#.