3. (Difficult) A geometric sequence with at least three terms and a rational common ratio has first term $32^{16}$ and last term $625^{30}$. If the product of all possible values of the second term of this sequence is $16^{x} \cdot 625^{y}$, where $x$ and $y$ are integers, what are the values of $x$ and $y$ ?
Answer: $x=214$ and $y=69$
Solution: Let $a$ be the common ratio of the geometric sequence and $k$ be the number of terms, then $32^{16} \cdot a^{k-1}=625^{30}$ or $2^{80} \cdot a^{k-1}=5^{120}$. Solving for $a$, we have
$$
a=\sqrt[k-1]{\frac{5^{120}}{2^{80}}}=\left(\frac{5^{120}}{2^{80}}\right)^{1 /(k-1)} .
$$

From the equation above, $k-1$ must divide both 80 and 120 . If $k-1$ is even, then we will consider the positive and negative values of the $(k-1)$ th root. Since the $\operatorname{GCF}(80,120)=40$, then the possible values of $k-1$ are $2,4,5,8,10,20,40$ (not 1 because there will only be two terms in the sequence). Thus, there are 13 possible values of $a$, which are
$$
\pm \frac{5^{60}}{2^{40}}, \pm \frac{5^{30}}{2^{20}}, \frac{5^{24}}{2^{16}}, \pm \frac{5^{15}}{2^{10}}, \pm \frac{5^{12}}{2^{8}}, \pm \frac{5^{6}}{2^{4}}, \pm \frac{5^{3}}{2^{2}}
$$

Thus, the product of all possible second terms of the sequence is
$$
\left(2^{80}\right)^{13}\left(\frac{5^{120+60+24+30+24+12+6}}{2^{80+40+16+20+16+8+4}}\right)=2^{1040}\left(\frac{5^{276}}{2^{184}}\right)=2^{856} \cdot 5^{276}=16^{214} \cdot 625^{69} .
$$

Therefore, $x=214$ and $y=69$.