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M21-37

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M21-37  [#permalink]

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New post 16 Sep 2014, 01:15
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E

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  95% (hard)

Question Stats:

35% (01:56) correct 65% (01:55) wrong based on 115 sessions

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Re M21-37  [#permalink]

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New post 16 Sep 2014, 01:15
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1
Official Solution:


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.


Answer: E
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Re: M21-37  [#permalink]

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New post 02 Jul 2015, 05:20
Hi,

Explanation seems good, but it look hard for me; could you develop with more examples or detail the answer?

Thanks a lot.

Luis Navarro
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Re: M21-37  [#permalink]

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New post 02 Jul 2015, 05:37
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luisnavarro wrote:
Hi,

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.
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Re: M21-37  [#permalink]

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New post 02 Jul 2015, 06:21
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Bunuel wrote:
luisnavarro wrote:
Hi,

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".

Thanks.

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Re: M21-37  [#permalink]

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New post 29 Jan 2019, 01:33
[quote="Bunuel"]Official Solution:


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.
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Re: M21-37  [#permalink]

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New post 29 Jan 2019, 01:37
1
Megha1119 wrote:
Bunuel wrote:
Official Solution:


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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Re: M21-37  [#permalink]

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New post 26 Feb 2019, 04:02
Thanks Bunuel. Makes sense now.

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Re: M21-37   [#permalink] 26 Feb 2019, 04:02
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