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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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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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

[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.

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

HI Bunuel:

I dont really understand the part where we need to consider 2^n.. Is it because 2^1/2 =1 ? or 2^0=1?

THank you !!!

In (1) we get that \(D[x] = 1\): 1 is the largest odd divisor of x. This implies that x is not divisible by 3, 5, 7, 9, ... So, x can only be 1 or 2, 4, 8, 16, ... (basically any power of 2: 2^1, 2^2, 2^3, 2^4, ...)
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