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If for any positive integer \(x\), \(d[x]\) denotes its smallest positive odd factor and \(D[x]\) denotes its largest odd factor, is \(x\) an even number?
(1) \(D[x] - d[x] = 0\)
(2) \(D[3x] = 3\)
(1) \(D[x] - d[x] = 0\)
(2) \(D[3x] = 3\)
16 Sep 2014, 01:15
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Official Solution:
If for any positive integer \(x\), \(d[x]\) denotes its smallest positive odd factor and \(D[x]\) denotes its largest odd factor, is \(x\) an even number?
Note that the smallest positive odd factor of any positive integer is 1, so \(d[x] = 1\) for any positive integer \(x\).
(1) \(D[x] - d[x] = 0\).
Since \(d[x]=1\), it follows that \(D[x] - 1 = 0\), thus \(D[x] = 1\). Furthermore, \(D[x] = 1\) implies that 1 is the largest odd factor of \(x\). As a result, \(x\) is not divisible by other odd numbers such as 3, 5, 7, 9, and so on. Therefore, \(x\) can only be 1 or any power of 2 (\(2^n\)), like 2, 4, 8, 16, etc. From this statement, we deduce that \(x\) can be either 1, which is odd, or \(2^n\), which is even. Not sufficient.
(2) \(D[3x] = 3\).
Again, \(x\) can be 1, which is odd, as the largest odd factor of \(3x=3\) is 3, or \(x\) can be \(2^n\) (like 2, 4, 8, etc.), making it even, as the largest odd factor of \(3*2^n\) is also 3 (for example, the largest odd factor of \(3*2=6\) or \(3*4=12\) is 3). Not sufficient.
(1)+(2) From both statements, \(x\) can be either 1 (odd) or \(2^n\) (even). Not sufficient.
Answer: E
If for any positive integer \(x\), \(d[x]\) denotes its smallest positive odd factor and \(D[x]\) denotes its largest odd factor, is \(x\) an even number?
Note that the smallest positive odd factor of any positive integer is 1, so \(d[x] = 1\) for any positive integer \(x\).
(1) \(D[x] - d[x] = 0\).
Since \(d[x]=1\), it follows that \(D[x] - 1 = 0\), thus \(D[x] = 1\). Furthermore, \(D[x] = 1\) implies that 1 is the largest odd factor of \(x\). As a result, \(x\) is not divisible by other odd numbers such as 3, 5, 7, 9, and so on. Therefore, \(x\) can only be 1 or any power of 2 (\(2^n\)), like 2, 4, 8, 16, etc. From this statement, we deduce that \(x\) can be either 1, which is odd, or \(2^n\), which is even. Not sufficient.
(2) \(D[3x] = 3\).
Again, \(x\) can be 1, which is odd, as the largest odd factor of \(3x=3\) is 3, or \(x\) can be \(2^n\) (like 2, 4, 8, etc.), making it even, as the largest odd factor of \(3*2^n\) is also 3 (for example, the largest odd factor of \(3*2=6\) or \(3*4=12\) is 3). Not sufficient.
(1)+(2) From both statements, \(x\) can be either 1 (odd) or \(2^n\) (even). Not sufficient.
Answer: E
General Discussion
