M07-02

Official Solution:


If \(L \not= 0\), is \(\frac{18K}{L}\) an integer?

It is crucial to remember that we are not given any information about whether K and L are integers. Keep this consideration in mind when evaluating the statements.

(1) \(\frac{K^2}{L^2}\) is an integer.

If \(K=L=1\), then \(\frac{18K}{L}=18\) and the answer to the question is YES. However, if \(K=\sqrt{2}\) and \(L=1\), then \(\frac{18K}{L}=18\sqrt{2}\), which is not an integer, and the answer to the question is NO. This statement is not sufficient.

Note that if we were told that K and L are integers, the first statement would have been sufficient since \(\frac{K^2}{L^2}\) being an integer would imply that \(\frac{K}{L}\) is an integer. This is because the ratio of two integers (\(\frac{K}{L}\)) can only be an integer or a non-integer rational number, but it cannot be an irrational number. However, \(\frac{K}{L}\) cannot be a non-integer rational number, because its square would not be an integer. Therefore, \(\frac{K}{L}\) must be an integer, making \(\frac{18K}{L}\) an integer.

(2) \(K - L = L\). \(K=2L\).

In this case, \(\frac{18K}{L}=\frac{18(2L)}{L}=36\), which is an integer. Thus, the answer to the question is YES. This statement is sufficient.


Answer: B