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# ­If x and y are integers and ((x - 1)(y - 1))! = 2, which of the

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Re: ­If x and y are integers and ((x - 1)(y - 1))! = 2, which of the [#permalink]
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Bunuel wrote:
­If $$x$$ and $$y$$ are integers and $$((x - 1)(y - 1))! = 2$$, which of the following could be a value of $$xy$$?

I. -6

II. 0

III. 6

A. I only
B. II only
C. II only
D. II and III only
E. I, II, and III­

$$((x - 1)(y - 1))! = 2$$

Therefore, we can conclude that

(x - 1)(y - 1) = 2

Case 1: (x - 1) = 2 & (y - 1) = 1 OR (x - 1) = 1 & (y - 1) = 2

xy = 6

Case 2: (x - 1) = -1 & (y - 1) = -2 OR (x - 1) = -2 & (y - 1) = -1

xy = 0

Option D
­
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Re: ­If x and y are integers and ((x - 1)(y - 1))! = 2, which of the [#permalink]
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Looking at $$­[(x-1)(y-1)]! = 2$$, one notices the factorial sign. The factorial which equals $$2$$ is $$2!$$. As $$x$$ and $$y$$ are integers, both brackets will be integers and thus will either both be positive or both negative. The only way two integers multiplied together is if the multiplying pairs are either $$2$$ and $$1$$ or $$-2$$ and $$-1$$.

Letting $$(x-1) = 2$$ and $$(y-1) = 1$$ gives that $$x = 3$$ and $$y = 2$$. Therefore $$xy = 6$$. Which is III.

Letting $$(x-1) = -2$$ and $$(y-1) = -1$$ gives that $$x = -1$$ and $$y = 0$$. Therefore $$xy = 0$$. Which is II.

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Re: ­If x and y are integers and ((x - 1)(y - 1))! = 2, which of the [#permalink]
Can x and y not be -3 and -2 as well ?
If we are trying to plug in values.This means xy can be 6 but also not equal to 2. Sorry if I am way off

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Re: ­If x and y are integers and ((x - 1)(y - 1))! = 2, which of the [#permalink]
K0315145 wrote:
Can x and y not be -3 and -2 as well ?
If we are trying to plug in values.This means xy can be 6 but also not equal to 2. Sorry if I am way off

Posted from my mobile deviceI'

I've posted a detailed solution HERE. Let me know if any questions.
Re: ­If x and y are integers and ((x - 1)(y - 1))! = 2, which of the [#permalink]
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