If x, y and z are positive integers, then...

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If x, y and z are positive integers, then...

by swerve » Wed Nov 01, 2017 9:48 am
If x, y and z are positive integers, then
$$x^2y^3z^4$$
must be divisible by which of the following?

$$I.\ \ x^2+y^3+z^4$$
$$II.\ \ xy+xz$$
$$III.\ \ xyz+z$$

A. None
B. I only
C. I and II only
D. I and III only
E. II and III only

The OA is A.

I don't understand this PS question. Please, can any expert help me to solve it? Thanks.

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by DavidG@VeritasPrep » Wed Nov 01, 2017 10:42 am
swerve wrote:If x, y and z are positive integers, then
$$x^2y^3z^4$$
must be divisible by which of the following?

$$I.\ \ x^2+y^3+z^4$$
$$II.\ \ xy+xz$$
$$III.\ \ xyz+z$$

A. None
B. I only
C. I and II only
D. I and III only
E. II and III only

The OA is A.

I don't understand this PS question. Please, can any expert help me to solve it? Thanks.
Take the easiest possible case: x = 1, y = 1, and z = 1.

$$x^2y^3z^4$$= 1.

$$I.\ \ x^2+y^3+z^4$$ --> 1 + 1 + 1 = 3. 1 is not divisible by 3. I is out.

$$II.\ \ xy+xz$$ --> 1 + 1 = 2. 1 is not divisible by 2. II is out.

$$III.\ \ xyz+z$$ --> 1 + 1 = 2. 1 is not divisible by 2. III is out.

So none of these statements needs to be true. The answer is A
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