8. The product of two natural numbers is 14000. What can be the maximum of their Greatest Common Divisor?
(A) 10
(B) 20
(C) 400
(D) 70
(E) 140
Problem 8. The answer is (B).

The prime factorization of 14000 is $2^{4} \cdot 5^{3} \cdot 7$. If the product of $a$ and $b$ is 14000, there will be a total of four factors of 2, three factors of 5, and one factor of 7 in the factorizations of $a$ and $b$. The 7 will therefore not appear in the factorization of the greatest common divisor, while the maximum number of factors of 2 and 5 will be achieved by placing two factors of 2 in each number, and dividing the three factors of 5 by placing one on one side and two on the other. The maximum value of the greatest common divisor is therefore $2^{2} \cdot 5=20$.

(Problem proposed by P. Francini)