Putnam 1998, Problem B-5

Let \( N \) be the positive integer with 1998 decimal digits, all of them 1; that is,
\[
N = 1111\cdots 11.
\]
Find the thousandth digit after the decimal point of \( \sqrt{N} \).

Solution:
Write \( N = \frac{10^{1998} - 1}{9} \). Then
\[
\sqrt{N} = \frac{10^{999}}{3} \sqrt{1 - 10^{-1998}} = \frac{10^{999}}{3} \left(1 - \frac{1}{2} 10^{-1998} + r\right),
\]
where \( r < 10^{-2000} \). Now the digits after the decimal point of \( 10^{999}/3 \) are given by \( .3333\ldots \), while the digits after the decimal point of \( \frac{1}{6} 10^{-999} \) are given by \( .00000\ldots 1666666\ldots \). It follows that the first 1000 digits of \( \sqrt{N} \) are given by \( .33333\ldots 3331 \); in particular, the thousandth digit is \( 1 \).