\section*{Problem 6 - 311046}

Let \(q\) be the larger of the two solutions to the equation \(x^{2}-4 x+1=0\).

Determine the last digit (units place) in the decimal digit representation of the number \(\left[q^{1992}\right]\).

Hint: If \(z\) is a real number, the integer \(g\) such that \(g \leq z < g+1\) is denoted by \(g = [z]\).
}

The solutions of the equation \(x^{2}-4 x+1=0\) are \(p=2-\sqrt{3}\) and \(q=2+\sqrt{3}\). For the numbers \(a_{n}=p^{n}+q^{n}\) \((n=0,1,2,3, \ldots)\), it holds that \(a_{n}>0\) as well as

\[
\begin{gathered}
a_{0}=2, \quad a_{1}=4, \quad a_{2}=4-4 \sqrt{3}+3+4+4 \sqrt{3}+3=14 \quad \text { and } \\
a_{n+2}-4 a_{n+1}+a_{n}=p^{n}\left(p^{2}-4 p+1\right)+q^{n}\left(a^{2}-4 q+1\right)=0
\end{gathered}
\]

By mathematical induction, it is now proven: For \(m=0,1,2, \ldots\), \(a_{3 m}, a_{3 m+1}, a_{3 m+2}\) are integers with 2, 4, and 4 as their last digits, respectively:

For \(m=0\), this follows from (1), and if it holds for some \(m=k \geq 0\), then according to (2), each of the following is

\[
a_{3(k+1)}=4 \cdot a_{3 k+2}-a_{3 k+1}
\]

an integer with the same last digit as \(4 \cdot 4-4=12\),

\[
a_{3(k+1)+1}=4 \cdot a_{3 k+3}-a_{3 k+2}
\]

an integer with the same last digit as \(4 \cdot 2-4=4\),

\[
a_{3(k+1)+2}=4 \cdot a_{3 k+4}-a_{3 k+3}
\]

an integer with the same last digit as \(4 \cdot 4-2=14\).

Since \(1992=3 \cdot 664\), \(a_{1992}\) has the last digit 2. Furthermore, \(0<p<1\), so \(p^{1992}-1<0<p^{1992}\). After adding \(q^{1992}\), it follows that \(a_{1992}-1<q^{1992}<a_{1992}\), i.e., \(q^{1992}\) lies between two integers whose last digits are 1 and 2, respectively. Thus, \(\left[q^{1992}\right]=1\).

Solution adopted from \([5]\)

\subsection*{7.34 XXXII. Olympiad 1992}

\subsection*{7.34.1 First Round 1992, Class 10}