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26 Sep 2019, 01:22
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
45% (01:43) correct
55%
(01:35)
wrong
based on 412
sessions
History
Date
Time
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Not Attempted Yet
How many values of Z satisfy \((|2Z|)! – |Z| = Z\) ?
A. 0
B. 1
C. 2
D. 3
E. More than 3
A. 0
B. 1
C. 2
D. 3
E. More than 3
22 Apr 2024, 00:46
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zxl2945
Bunuel
There are discussions about Z taking the value 1, 1/2
Could you please kindly explain why can't Z take the value 0?
Thank you
How many values of Z satisfy \((|2Z|)! – |Z| = Z\) ?There are discussions about Z taking the value 1, 1/2
Could you please kindly explain why can't Z take the value 0?
Thank you
A. 0
B. 1
C. 2
D. 3
E. More than 3
Let's check two cases for \((|2z|)! - |z| = z\)
If \(z < 0\), then we'd have:
\((-2z)! - (-z) = z\)
\((-2z)! = 0\)
If \(z ≥ 0\), then we'd have:
\((2z)! - z = z\)
\((2z)! = 2z\)
Answer: C.
P.S. z cannot be 0 because this value does not satisfy \((|2z|)! - |z| = z\). If \(z = 0\), we get:
\((|0|)! - |0| = 0\)
\(0! - 0 = 0\)
\(1 - 0 = 0\) (recall that \(0! = 1\))
\(1 = 0\)
General Discussion
26 Sep 2019, 03:31
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Z is obviously positive since |(2Z)|! is always greater than |z|. This means the equation can be rewritten as 2Z!-Z=Z
this gives us 2Z!=2Z; This also can be expanded as 2Z(2Z-1)!=2Z.
This gives us (2Z-1)!=1.
There are only two numbers whose factorials equal 1- that is 1 itself and 0.
2Z-1=0;Z=1/2
2Z-1=1;z=1
Put them both back into the equation to ascertain that they hold and they do
hence there are 2 distinct values of z.
Answer is C
this gives us 2Z!=2Z; This also can be expanded as 2Z(2Z-1)!=2Z.
This gives us (2Z-1)!=1.
There are only two numbers whose factorials equal 1- that is 1 itself and 0.
2Z-1=0;Z=1/2
2Z-1=1;z=1
Put them both back into the equation to ascertain that they hold and they do
hence there are 2 distinct values of z.
Answer is C
