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16 Sep 2014, 01:26
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If \(x\), \(y\), and \(z\) are positive integers and \(xyz=2,700\). Is \(\sqrt{x}\) an integer?
(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube.
(2) \(\sqrt{z}\) is not an integer.
(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube.
(2) \(\sqrt{z}\) is not an integer.
16 Sep 2014, 01:26
Kudos
Bookmarks
Official Solution:
If \(x\), \(y\), and \(z\) are positive integers and \(xyz=2,700\). Is \(\sqrt{x}\) an integer?
Note: a perfect square is an integer that can be written as the square of an integer. For example, \(16=4^2\) is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of an integer. For example, \(27=3^3\) is a perfect cube.
First, find the prime factorization of 2,700: \(xyz=2,700=2^2*3^3*5^2\).
Now, let's examine the given statements:
(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube.
Since \(z\) can take fewer values than \(y\), for simplicity, let's analyze it from \(z\) perspective. Since given that \(z\) is an odd perfect cube, it can be \(1=1^3\) or \(3^3\).
If \(z = 3^3\), \(y\) can only be \(2^2\) or \(2^2*5^2\). Respectively, \(x\) will be \(5^2\) or \(1\). For each of these values, \(\sqrt{x}\) is an integer.
If \(z = 1\), \(y\) can be \(2^2\), \(2^2*3^2\), \(2^2*5^2\), or \(2^2*3^2*5^2\). For the first three cases, \(x\) turns out NOT to be a perfect square (\(3^3*5^2\), \(3*5^2\), and \(3^3\)). However, for the fourth case, \(\sqrt{x}\) is a perfect square (\(1\)).
Not sufficient.
(2) \(\sqrt{z}\) is not an integer. This statement is clearly insufficient, as it does not provide enough information about \(x\) and \(y\).
(1)+(2) Since from (2) \(\sqrt{z} \ne integer\), then \(z \ne 1\), and thus from (1) it can only be \(3^3\). As discussed above, if \(z=3^3\), then \(x\) will be \(5^2\) or \(1\). For each of these values, \(\sqrt{x}\) is an integer. Sufficient.
Answer: C
If \(x\), \(y\), and \(z\) are positive integers and \(xyz=2,700\). Is \(\sqrt{x}\) an integer?
Note: a perfect square is an integer that can be written as the square of an integer. For example, \(16=4^2\) is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of an integer. For example, \(27=3^3\) is a perfect cube.
First, find the prime factorization of 2,700: \(xyz=2,700=2^2*3^3*5^2\).
Now, let's examine the given statements:
(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube.
Since \(z\) can take fewer values than \(y\), for simplicity, let's analyze it from \(z\) perspective. Since given that \(z\) is an odd perfect cube, it can be \(1=1^3\) or \(3^3\).
If \(z = 3^3\), \(y\) can only be \(2^2\) or \(2^2*5^2\). Respectively, \(x\) will be \(5^2\) or \(1\). For each of these values, \(\sqrt{x}\) is an integer.
If \(z = 1\), \(y\) can be \(2^2\), \(2^2*3^2\), \(2^2*5^2\), or \(2^2*3^2*5^2\). For the first three cases, \(x\) turns out NOT to be a perfect square (\(3^3*5^2\), \(3*5^2\), and \(3^3\)). However, for the fourth case, \(\sqrt{x}\) is a perfect square (\(1\)).
Not sufficient.
(2) \(\sqrt{z}\) is not an integer. This statement is clearly insufficient, as it does not provide enough information about \(x\) and \(y\).
(1)+(2) Since from (2) \(\sqrt{z} \ne integer\), then \(z \ne 1\), and thus from (1) it can only be \(3^3\). As discussed above, if \(z=3^3\), then \(x\) will be \(5^2\) or \(1\). For each of these values, \(\sqrt{x}\) is an integer. Sufficient.
Answer: C
General Discussion
