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Bunuel
If \(z\) is a positive integer, is 2 a factor of \(z\)?


(1) 22 is a factor of \(15z\)

(2) 22 is a factor of \(16z\)


(1) 22 is a factor of \(15z\)

22=2*11
15z=3*5*z

Since there is nothing common between known factors of the two numbers, It can be inferred that z has to have at least a 2 and a 11.

Sufficient

Answer is either A or D.

(2) 22 is a factor of \(16z\)

22=2*11
16z=(2^4)*z

Since 2 is common we cannot infer whether z has a 2 or not. We can be sure that z has at least one 11 because that is not common.

Not Sufficient.

Answer (A)
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Good question.
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Bunuel
If \(z\) is a positive integer, is 2 a factor of \(z\)?


(1) 22 is a factor of \(15z\)

(2) 22 is a factor of \(16z\)

(1) As 15 is an odd number so z must be an even number and is divisible by 2. Sufficient.

(2) As 16 is even then z can be odd or can be even, for example; z is 11 the 16z is divisible by 2 but z is not divisible 2, but z can be 22 and is divisible by 2. Yes and No. Insufficient.

The answer is A
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My take on this question

is \(2\) a factor of \(Z\)?

(1):
\(22\) is a factor of \(15Z\)
or in other words

\(\frac{(3 * 5 *Z)}{(2 * 11)} = integer\)

since \(15\) is not a multiple of 2 nor 11, thus \(Z\) must be a multiple of \(2\) and a multiple of \(11\), otherwise the premise doesn't hold
Sufficient

(2):
\(22\) is a factor of \(16Z\)


or in other words

\(\frac{(2^4 *Z)}{(2 * 11)} = integer\)
\(\frac{(2^3 *Z)}{(11)} = integer\)

now \(Z\) does not be a multiple of \(2\) for this to hold, for example:
\(Z = 11\) then the result is \(2^3\) is an integer => statement is FALSE
\(Z = 22\) then the result is \(2^3 * 2\) is also an integer => statement is TRUE
Insufficient
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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