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Re: What is the value of Z if (|2Z|)! – |Z| = Z ? [#permalink]
itoyj wrote:
Bunuel
Can u please help
The given equation is satisfied by
1,1/2

Posted from my mobile device



How many values of Z satisfy (|2Z|)!–|Z|=Z(|2Z|)!–|Z|=Z ?

Z cannot be negative no since

(|2Z|)! = Z + IZI and if Z is negative then RHS is 0 but factorial never be 0.
Aso if z=3 then (|2Z|)! = 720 ..far larger than RHS value
Check if z=2 works then LHS = 4*3*2= 24 ,again far larger value than 4 (RHS part)
now we see for (|2Z|)! to be integeger only Z= 1 & 1/2 works
Check both and get the desired answers

Answer C
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Re: What is the value of Z if (|2Z|)! |Z| = Z ? [#permalink]
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
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What is the value of Z if (|2Z|)! |Z| = Z ? [#permalink]
Expert Reply
 
zxl2945 wrote:
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 \((|2Z|)! – |Z| = Z\) ?

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:

\((-2z)! - (-z) = z\)
\((-2z)! = 0\)

Since the factorial of none of the numbers equals 0, then this equation has no solution. Thus, \((|2z|)! - |z| = z\) does not have a solution when \(z < 0\).

If \(z ≥ 0\), then we'd have:

\((2z)! - z = z\)
\((2z)! = 2z\)

For integer n, \(n! = n\) only in two cases: when \(n = 1\) and when \(n = 2\), in all other cases \(n! > n\). Thus, either \(2z = 1\) or \(2z = 2\). Therefore, \(z = \frac{1}{2}\) or \(z = 1\).

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 \(0! = 1\))

\(1 = 0\)

Which is not true.

 ­
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Re: What is the value of Z if (|2Z|)! |Z| = Z ? [#permalink]
Bunuel wrote:
zxl2945 wrote:
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 \((|2Z|)! – |Z| = Z\) ?

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:



\((-2z)! - (-z) = z\)
\((-2z)! = 0\)

Since the factorial of none of the numbers equals 0, then this equation has no solution. Thus, \((|2z|)! - |z| = z\) does not have a solution when \(z < 0\).

If \(z ≥ 0\), then we'd have:



\((2z)! - z = z\)
\((2z)! = 2z\)

For integer n, \(n! = n\) only in two cases: when \(n = 1\) and when \(n = 2\), in all other cases \(n! > n\). Thus, either \(2z = 1\) or \(2z = 2\). Therefore, \(z = \frac{1}{2}\) or \(z = 1\).

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 \(0! = 1\))

\(1 = 0\)

Which is not true.

 ­

­Thank you for your kind reply! I finally understand.­
GMAT Club Bot
Re: What is the value of Z if (|2Z|)! |Z| = Z ? [#permalink]
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