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16 Sep 2014, 00:14
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If \(z\) is a non-negative integer, is \(\frac{75}{z+1}\) an integer?
(1) \(z\) is a multiple of 75
(2) \(0 \le z \le 75\)
(1) \(z\) is a multiple of 75
(2) \(0 \le z \le 75\)
16 Sep 2014, 00:14
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Official Solution:
If \(z\) is a non-negative integer, is \(\frac{75}{z+1}\) an integer?
(1) \(z\) is a multiple of 75.
Since \(z\) is a non-negative integer, the above implies that \(z\) can be 0, 75, 150, and so on. If \(z = 0\), then \(\frac{75}{z+1}=75\), which is an integer. However, if \(z\) is any other number, then \(\frac{75}{z+1}\) will not be an integer. Not sufficient.
(2) \(0 \le z \le 75\).
The above is clearly insufficient.
(1)+(2) Since from (2) \(0 \le z \le 75\), then \(z\) can only be 0 or 75. If \(z=0\), then \(\frac{75}{z+1}=75\), which is an integer. However, if \(z=75\), then \(\frac{75}{z+1}=\frac{75}{76}\), which is not an integer. Not sufficient.
Answer: E
If \(z\) is a non-negative integer, is \(\frac{75}{z+1}\) an integer?
(1) \(z\) is a multiple of 75.
Since \(z\) is a non-negative integer, the above implies that \(z\) can be 0, 75, 150, and so on. If \(z = 0\), then \(\frac{75}{z+1}=75\), which is an integer. However, if \(z\) is any other number, then \(\frac{75}{z+1}\) will not be an integer. Not sufficient.
(2) \(0 \le z \le 75\).
The above is clearly insufficient.
(1)+(2) Since from (2) \(0 \le z \le 75\), then \(z\) can only be 0 or 75. If \(z=0\), then \(\frac{75}{z+1}=75\), which is an integer. However, if \(z=75\), then \(\frac{75}{z+1}=\frac{75}{76}\), which is not an integer. Not sufficient.
Answer: E
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30 May 2016, 09:46
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I think this is a high-quality question and I agree with explanation.
