# Difference between revisions of "1990 AHSME Problems/Problem 30"

## Problem

If $R_n=\tfrac{1}{2}(a^n+b^n)$ where $a=3+2\sqrt{2}$ and $b=3-2\sqrt{2}$, and $n=0,1,2,\cdots,$ then $R_{12345}$ is an integer. Its units digit is

$\text{(A) } 1\quad \text{(B) } 3\quad \text{(C) } 5\quad \text{(D) } 7\quad \text{(E) } 9$

## Solution

$(a+b)R_n=\tfrac12(a^{n+1}+b^{n+1})+ab\cdot\tfrac12(a^{n-1}+b^{n-1})=R_{n+1}+abR_{n-1}$ but $a+b=6$ and $ab=1$, so this means that $R_{n+1}=6R_n-R_{n-1}$. Since $R_0=1$ and $R_1=3$, all terms are integers and we can continue the recurrence $\rm{mod}\ 10$ to get the repeating sequence $1,3,7,9,7,3$. The number $12345$ is divisible by three but not by six, so the required element in the sequence is $9$ which is answer $\fbox{E}$