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If for any positive integer x, d[x] denotes its smallest odd  [#permalink]

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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?

(1) D[x] - d[x] = 0
(2) D[3x] = 3

Originally posted by reg123456 on 01 Nov 2010, 08:58.
Last edited by Bunuel on 12 Jul 2013, 02:13, edited 1 time in total.
Edited the question and added the OA
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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.

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Re: gmat club test DS - special operations  [#permalink]

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x = 1 and 2 passes the conditions in both a and b.

thus E
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Re: If for any positive integer x, d[x] denotes its smallest odd  [#permalink]

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Bunuel wrote:
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.

Hello Bunuel, question mentions that x is positive integer but did not mention d[x] to be smallest positive odd divisor right?! how can you consider d[x] to be 1? I want to consider d[x] as negative of highest positive odd divisor!
However, answer to the question will remain the same E though.
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Re: If for any positive integer x, d[x] denotes its smallest odd  [#permalink]

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nitinj025 wrote:
HI,
For the First Condition to Satisfy.
i)D[x]-d[x] =0 the value should be a prime number . Which, directly states that value is not even.
ii) D[3x] =1 the value can only be 1 for x. Which even states that the value is not even.
Please, Can you explain my doubt.
Regards,
NJ

Hi NJ,

1. D[x] - d[x] = 0 implies that the smallest odd divisor and largest odd divisor are the same. This divisor can also be 1. i.e. not even.
For x = 1, D[x] = 1 and d[x] = 1 => x is not even
For x = 2, D[x] = 1 and d[x] = 1 => x is even
Insufficient.

You have incorrectly read the statement 2 to be D[3x] = 1.
2. D[3x] = 3
For x = 1, D[3x] = 3, x is not even
For x = 2, D[3x] = 3 (because 3x = 6, factors of 6 = 1, 2, 3, 6 i.e. largest odd factor =3) i.e. x is even.
For x = 3, D[3x] = 9 (because 3x = 9, factors of 6 = 1, 3, 9 i.e. largest odd factor =9) - Not relevant as it does not satisfy D[3x] = 3.
Insufficient.

Together, based on statement 2, only possible values for x are 1 and 2.
x = 1, d[x] = 1, D[x] = 1, D[3x] = 3, x is odd
x = 2, d[x] = 1, D[x] = 1, D[3x] = 1, x is even
Therefore, together insufficient. i.e Answer is E.

Hope this helps.
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Re: If for any positive integer x, d[x] denotes its smallest odd  [#permalink]

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

There is one variable (x) and 2 more equations are given by the 2 conditions.
For condition 1, D[x]=d[x], the answer is 'yes' for x=1, but 'no' for x=2. This is insufficient.
For condition 2, D[3x]=3, this also gives 'yes' for x=1, but 'no' for x=2. This is insufficient.
Looking at them together, it still gives 'yes' for x=1, but 'no' for x=2. The answer is therefore (E).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If for any positive integer x, d[x] denotes its smallest odd  [#permalink]

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Bunuel, how can x be 2^n?
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Re: If for any positive integer x, d[x] denotes its smallest odd  [#permalink]

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mahrah wrote:
Bunuel, how can x be 2^n?

D[x] denotes the largest odd divisor of x. (1) says that D[x] = 1, so the the largest odd divisor of x. This means that x is some power of 2: 2^0 = 1, 2^1 = 2, 2^2 = 4, ... The largest, and the only, odd divisor of all of those numbers is 1.
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Re: If for any positive integer x, d[x] denotes its smallest odd  [#permalink]

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_________________ Re: If for any positive integer x, d[x] denotes its smallest odd   [#permalink] 13 Apr 2019, 09:36
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