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Math Expert V
Joined: 02 Sep 2009
Posts: 58434

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Difficulty:   95% (hard)

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

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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?

(1) $$D[x] - d[x] = 0$$

(2) $$D[3x] = 3$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 58434

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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.

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

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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.

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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.

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.

Luis Navarro
Looking for 700
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Joined: 29 Jun 2017
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[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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

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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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Thanks Bunuel. Makes sense now.

Posted from my mobile device Re: M21-37   [#permalink] 26 Feb 2019, 04:02
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