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If \(N = 1234x\), where \(x\) is the units digit of \(N\), is \(N\) a multiple of 5?
(1) \(x!\) is not divisible by 5
(2) \(x\) is divisible by 9
(1) \(x!\) is not divisible by 5
(2) \(x\) is divisible by 9
16 Sep 2014, 00:11
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Official Solution:
If \(N = 1234x\), where \(x\) is the units digit of \(N\), is \(N\) a multiple of 5?
In order for \(1234x\) to be divisible by 5, \(x\) must be either 0 or 5. So, the question asks whether \(x\) equals to 0 or 5.
(1) \(x!\) is not divisible by 5.
Since \(x\) is a digit, \(x\) can be 0, 1, 2, 3, or 4 (note that 0!=1). If \(x=0\), \(N\) will be divisible by 5 but if \(x\) is1, 2, 3, or 4, \(N\) will not be divisible by 5. Not sufficient.
(2) \(x\) is divisible by 9.
Since \(x\) is a digit, \(x\) can be 0 or 9 (note that zero is divisible by every integer except zero itself). Not sufficient.
(1)+(2) The only value that satisfies both statements is \(x=0\). Thus, \(N = 12340\), which is divisible by 5. Sufficient.
Answer: C
If \(N = 1234x\), where \(x\) is the units digit of \(N\), is \(N\) a multiple of 5?
In order for \(1234x\) to be divisible by 5, \(x\) must be either 0 or 5. So, the question asks whether \(x\) equals to 0 or 5.
(1) \(x!\) is not divisible by 5.
Since \(x\) is a digit, \(x\) can be 0, 1, 2, 3, or 4 (note that 0!=1). If \(x=0\), \(N\) will be divisible by 5 but if \(x\) is1, 2, 3, or 4, \(N\) will not be divisible by 5. Not sufficient.
(2) \(x\) is divisible by 9.
Since \(x\) is a digit, \(x\) can be 0 or 9 (note that zero is divisible by every integer except zero itself). Not sufficient.
(1)+(2) The only value that satisfies both statements is \(x=0\). Thus, \(N = 12340\), which is divisible by 5. Sufficient.
Answer: C
General Discussion
03 Mar 2023, 15:40
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
