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If for any positive integer \(x\), \(d[x]\) denotes its smallest positive 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 positive integer \(x\).
(1) \(D[x] - d[x] = 0\). Since \(d[x]=1\), then \(D[x] - 1 = 0\), which gives \(D[x] = 1\). \(x\) 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.
Explanation seems good, but it look hard for me; could you develop with more examples or detail the answer?
Thanks a lot.
Luis Navarro Looking for 700
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.
Explanation seems good, but it look hard for me; could you develop with more examples or detail the answer?
Thanks a lot.
Luis Navarro Looking for 700
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.
Very interesting, great¡¡¡ Actually in my analysis I had not view the 2^n posibilities, and now it is clear to me: For all 2^n the smallest and greatest odd divisor are the same = 1. It is easy to understand when you actually see explanations like yours but, its hard to do "alone".
If for any positive integer \(x\), \(d[x]\) denotes its smallest positive 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 positive integer \(x\).
(1) \(D[x] - d[x] = 0\). Since \(d[x]=1\), then \(D[x] - 1 = 0\), which gives \(D[x] = 1\). \(x\) 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.
Hi Bunuel, Can you please explain why we need to check for 2^n? Doesn't the problem end at x=1?
If for any positive integer \(x\), \(d[x]\) denotes its smallest positive 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 positive integer \(x\).
(1) \(D[x] - d[x] = 0\). Since \(d[x]=1\), then \(D[x] - 1 = 0\), which gives \(D[x] = 1\). \(x\) 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.
Hi Bunuel, Can you please explain why we need to check for 2^n? Doesn't the problem end at x=1?
Thanks, Megha.
No. x can be 1 or 2, 4, 8, 16, 32, ... Notice that for all those numbers both statements hold true. For example, for all of them both smallest and largest odd divisors is 1.
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35% (01:56) correct 
