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16 Sep 2014, 00:34
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If \(L \not= 0\), is \(\frac{18K}{L}\) an integer?
(1) \(\frac{K^2}{L^2}\) is an integer.
(2) \(K - L = L\)
(1) \(\frac{K^2}{L^2}\) is an integer.
(2) \(K - L = L\)
16 Sep 2014, 00:34
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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
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
General Discussion
JackSparr0w
22 Nov 2014, 14:18
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Curious to see what other people's first reaction to seeing this is.
My first thought was: for (18k)/L to be an integer, then L must be a factor of either 18, or K, or both. I thought stmt 1 was sufficient to determine that L was a factor of K but I guess this is incorrect.
I'm assuming it is always the case that (x^2)/(y^2) will never be sufficient to determine that y is a factor of x.
Stmt 2 tells us that L is a factor of K (because K=2L) therefore we know the L in the denominator will cancel out one factor of L in the numerator. This leaves a 1 in the denominator as well.
My first thought was: for (18k)/L to be an integer, then L must be a factor of either 18, or K, or both. I thought stmt 1 was sufficient to determine that L was a factor of K but I guess this is incorrect.
I'm assuming it is always the case that (x^2)/(y^2) will never be sufficient to determine that y is a factor of x.
Stmt 2 tells us that L is a factor of K (because K=2L) therefore we know the L in the denominator will cancel out one factor of L in the numerator. This leaves a 1 in the denominator as well.
