Problem (Number Theory):
Let $T=\sqrt{26}$. Compute the number of ordered pairs of positive integers $(a, b)$ with the property that $a b=T^{20} \cdot 210^{12}$, and the greatest common divisor of $a$ and $b$ is 1 .

Solution:
If the prime factorization of $a b$ is $p_{1}^{e_{1}} p_{2}^{e_{2}} \ldots p_{k}^{e_{k}}$, where the $p_{i}$ 's are distinct primes and the $e_{i}$ 's are positive integers, then in order for $\operatorname{gcd}(a, b)$ to equal 1 , each $p_{i}$ must be a divisor of exactly one of $a$ or $b$. Thus the desired number of ordered pairs is $2^{k}$ because there are 2 choices for each prime divisor (i.e., $p_{i} \mid a$ or $p_{i} \mid b$ ). With $T=\sqrt{26}$, it follows that $(\sqrt{26})^{20} \cdot 210^{12}=\left(2^{10} \cdot 13^{10}\right) \cdot 210^{12}=2^{22} \cdot 3^{12} \cdot 5^{12} \cdot 7^{12} \cdot 13^{10}$. Thus there are five distinct prime divisors, and the answer is $2^{5}=\mathbf{3 2}$.

Final answer: 32