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16 Sep 2014, 00:59
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Is \(10xy\) an even integer?
(1) \(7 + x + y\) is an even integer
(2) \(x - y\) is an odd integer
(1) \(7 + x + y\) is an even integer
(2) \(x - y\) is an odd integer
16 Sep 2014, 00:59
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Official Solution:
Is \(10xy\) an even integer?
In order for \(10xy=2*5xy\) to be even, \(5xy\) must be an integer. In this case, \(10xy=2*5xy\) would be \(2*integer\), which is an even number. Hence, essentially, the question is asking whether \(5xy\) is an integer.
However, it's important to note that \(x\) and \(y\) themselves don't necessarily have to be integers for \(5xy\) to be an integer. For instance, if \(x=\frac{7}{9}\) and \(y=\frac{9}{7}\), \(5xy\) is an integer. Nonetheless, if both \(x\) and \(y\) are integers, then \(5xy\) is certainly an integer, thereby making \(10xy\) even.
(1) \(7 + x + y\) is an even integer.
Since \(7+x+y\) is even, then \(x+y\) must be odd. If \(x=1\) and \(y=2\), the answer will be YES. However, if \(x=1.3\) and \(y=1.7\), the answer will be NO. Not sufficient.
(2) \(x-y\) is an odd integer.
If \(x=1\) and \(y=2\), the answer will be YES. However, if \(x=1.3\) and \(y=0.3\), the answer will be NO. Not sufficient.
(1)+(2) From (1) we know that \(x+y=odd\) and from (2) we know that \(x-y=odd\). Adding these two equations gives \((x+y)+(x-y)=odd_1+odd_2\), which results in \(2x=even\). This implies that \(x=\frac{even}{2}\), which is an integer, and thus \(y\) is also an integer. Therefore, \(10xy=10*integer=even\). Sufficient.
Answer: C
Is \(10xy\) an even integer?
In order for \(10xy=2*5xy\) to be even, \(5xy\) must be an integer. In this case, \(10xy=2*5xy\) would be \(2*integer\), which is an even number. Hence, essentially, the question is asking whether \(5xy\) is an integer.
However, it's important to note that \(x\) and \(y\) themselves don't necessarily have to be integers for \(5xy\) to be an integer. For instance, if \(x=\frac{7}{9}\) and \(y=\frac{9}{7}\), \(5xy\) is an integer. Nonetheless, if both \(x\) and \(y\) are integers, then \(5xy\) is certainly an integer, thereby making \(10xy\) even.
(1) \(7 + x + y\) is an even integer.
Since \(7+x+y\) is even, then \(x+y\) must be odd. If \(x=1\) and \(y=2\), the answer will be YES. However, if \(x=1.3\) and \(y=1.7\), the answer will be NO. Not sufficient.
(2) \(x-y\) is an odd integer.
If \(x=1\) and \(y=2\), the answer will be YES. However, if \(x=1.3\) and \(y=0.3\), the answer will be NO. Not sufficient.
(1)+(2) From (1) we know that \(x+y=odd\) and from (2) we know that \(x-y=odd\). Adding these two equations gives \((x+y)+(x-y)=odd_1+odd_2\), which results in \(2x=even\). This implies that \(x=\frac{even}{2}\), which is an integer, and thus \(y\) is also an integer. Therefore, \(10xy=10*integer=even\). Sufficient.
Answer: C
General Discussion
18 Mar 2016, 03:37
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Hi,
Hope zero can be considered rite?
If its taken, then the answer is E. Kindly debrief?
Hope zero can be considered rite?
If its taken, then the answer is E. Kindly debrief?
