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# What is the value of Z if (|2Z|)! – |Z| = Z ?

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

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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]

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 $$z = 1$$.

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 $$z = 1$$.

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.

­