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The positive integers x, y, and z are such that x is a factor of y and

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abhimahna

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15 Mar 2018, 11:54

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dave13

If x, y, and z are 3, 9 and 27 repectivey then how can A be sufficient ?

Hey dave13 ,

Statement 1 says xz is even. That means atleast one of them has to be even. But unfortunately, in the examples you took, both x and z are odd. Hence, invalid values.

Does that make sense?

dave13

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15 Mar 2018, 12:02

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abhimahna

dave13

If x, y, and z are 3, 9 and 27 repectivey then how can A be sufficient ?

YaY! Thank you! yes it does awesome sense of crystal logic

and why i make things complicated

it seems I dont follow DS rules...

Eduardo86

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07 Aug 2020, 09:31

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Bunuel

amitgovin

The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?

(1) xz is even

(2) y is even.

Given:
x is a factor of y --> \(y=mx\), for some non-zero integer \(m\);
y is a factor of z --> \(z=ny\), for some non-zero integer \(n\);
So, \(z=mnx\).

Question: is z even? Note that \(z\) will be even if either \(x\) or \(y\) is even

(1) \(xz\) even --> either \(z\) even, so the answer is directly YES or \(x\) is even (or both). But if \(x\) is even and as \(z=mnx\) then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.

Never mind i get it. If x is even y will be even and so z will be even too.
2) y is even so z is even.

Forgot to join the information with the info of the question
(2) \(y\) even --> as \(z=ny\) then as one of the multiples of z even --> z even. Sufficient.

Answer: D.

Why z cannot be 3 or another odd nomrber for the 1 statement z=3 and x=2 will be even too.

Bunuel

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07 Aug 2020, 09:37

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Eduardo86

Bunuel

amitgovin

The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?

(1) xz is even

(2) y is even.

Given:
x is a factor of y --> \(y=mx\), for some non-zero integer \(m\);
y is a factor of z --> \(z=ny\), for some non-zero integer \(n\);
So, \(z=mnx\).

Question: is z even? Note that \(z\) will be even if either \(x\) or \(y\) is even

(1) \(xz\) even --> either \(z\) even, so the answer is directly YES or \(x\) is even (or both). But if \(x\) is even and as \(z=mnx\) then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.

(2) \(y\) even --> as \(z=ny\) then as one of the multiples of z even --> z even. Sufficient.

Answer: D.

Why z cannot be 3 or another odd nomrber for the 1 statement z=3 and x=2 will be even too.

x must be a factor of z. If z=3 and x=2, the x is not a factor of z.

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