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)#?
1 Answer
Jun 29, 2017
Explanation:
Note that the given integer is
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