16. Let $a_{1}=21$ and $a_{2}=90$, and for $n \geq 3$, let $a_{n}$ be the last two digits of $a_{n-1}+a_{n-2}$. What is the remainder of $a_{1}^{2}+a_{2}^{2}+\cdots+a_{20}^{2}$ when it is divided by 8 .
16. Ans: 1
Note that the value of $x^{2}(\bmod 8)$ is completely determined by the value of $x$ $(\bmod 4)$. Furthermore, $a_{n} \equiv a_{n-1}+a_{n-2} \quad(\bmod 100) \equiv a_{n-1}+a_{n-2}(\bmod 4)$. Thus the sequence $\left(a_{n}(\bmod 4)\right)_{n \in Z}+$ equals $(1,2,3,1,0,1,1,2,3,1,0,1,1,2 \ldots)$, which has a cycle of length 6. Also, the sum of the squares in one cycle is 0 $(\bmod 8)$. Now $2005=6 \times 334+1$. Thus the required answer is 1 .