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

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.

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.

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.

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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