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If x and y are integers and ((x - 1)(y - 1))! = 2, which of the
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29 Apr 2024, 01:30
29 Apr 2024, 01:30
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Difficulty:
55%
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Question Stats:
65% (02:18) correct
35%
(02:09)
wrong
based on 107
sessions
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
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
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Re: If x and y are integers and ((x - 1)(y - 1))! = 2, which of the
[#permalink]
05 May 2024, 08:15
05 May 2024, 08:15
Expert Reply
Official Solution:
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
Only \(2! = 2\), so we have that \((x - 1)(y - 1) = 2\).
Since \(x\) and \(y\) are integers, we can have the following four cases:
\((x - 1)=2\) and \((y - 1) = 1\). This gives \(x=3\) and \(y=2\).
\((x - 1)=1\) and \((y - 1) = 2\). This gives \(x=2\) and \(y=3\).
\((x - 1)=-2\) and \((y - 1) = -1\). This gives \(x=-1\) and \(y=0\).
\((x - 1)=-1\) and \((y - 1) = -2\). This gives \(x=0\) and \(y=-1\).
The first two cases yield \(xy=6\), and the last two cases yield \(xy=0\).
Answer: D
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
Only \(2! = 2\), so we have that \((x - 1)(y - 1) = 2\).
Since \(x\) and \(y\) are integers, we can have the following four cases:
\((x - 1)=2\) and \((y - 1) = 1\). This gives \(x=3\) and \(y=2\).
\((x - 1)=1\) and \((y - 1) = 2\). This gives \(x=2\) and \(y=3\).
\((x - 1)=-2\) and \((y - 1) = -1\). This gives \(x=-1\) and \(y=0\).
\((x - 1)=-1\) and \((y - 1) = -2\). This gives \(x=0\) and \(y=-1\).
The first two cases yield \(xy=6\), and the last two cases yield \(xy=0\).
Answer: D
General Discussion
Re: If x and y are integers and ((x - 1)(y - 1))! = 2, which of the
[#permalink]
29 Apr 2024, 01:57
29 Apr 2024, 01:57
1
Kudos
Only 2! = 2 so (x-1) * (y-1) = 2 ; x-1 = 2 or y-1= 1 or x =3 and y= 2. Alternatively they could be -2 and -1 in which case x=-1 and y=0 so 6 and 0 are two possible options.
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