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E
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Difficulty:
95%
(hard)
Question Stats:
32% (02:10) correct
68%
(01:58)
wrong
based on 479
sessions
History
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Not Attempted Yet
If for any positive integer x, d[x] denotes its smallest odd divisor and D[x] denotes its largest odd divisor, is x even?
(1) D[x] - d[x] = 0
(2) D[3x] = 3
(1) D[x] - d[x] = 0
(2) D[3x] = 3
![If for any positive integer x, d[x] denotes its smallest odd](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If for any positive integer x, d[x] denotes its smallest odd"){kind=icon}
01 Nov 2010, 09:13
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If for any positive integer x, d[x] denotes its smallest odd divisor and D[x] denotes its largest odd divisor, is x even?
First of all note that the smallest positive odd divisor of any positive integer is 1. Thus \(d[x]=1\) for any x.
(1) D[x] - d[x] = 0 --> \(D[x] - 1 = 0\) --> \(D[x] = 1\) --> x can be 1, so odd or \(2^n\), (2, 4, 8, ...), so even. Not sufficient.
(2) D[3x] = 3 --> again x can be 1, so odd, as the largest odd divisor of \(3x=3\) is 3 or x can be \(2^n\) (2, 4, 8, ...), so even, as the largest odd divisor of 3*2=6 or 2*4=12 is 3. Not sufficient.
(1)+(2) From (1) and (2) we have that x can be either 1, so odd or 2^n, so even. Not sufficient.
Answer: E.
First of all note that the smallest positive odd divisor of any positive integer is 1. Thus \(d[x]=1\) for any x.
(1) D[x] - d[x] = 0 --> \(D[x] - 1 = 0\) --> \(D[x] = 1\) --> x can be 1, so odd or \(2^n\), (2, 4, 8, ...), so even. Not sufficient.
(2) D[3x] = 3 --> again x can be 1, so odd, as the largest odd divisor of \(3x=3\) is 3 or x can be \(2^n\) (2, 4, 8, ...), so even, as the largest odd divisor of 3*2=6 or 2*4=12 is 3. Not sufficient.
(1)+(2) From (1) and (2) we have that x can be either 1, so odd or 2^n, so even. Not sufficient.
Answer: E.
General Discussion
05 May 2011, 22:49
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x = 1 and 2 passes the conditions in both a and b.
thus E
thus E
