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# PS: FREAK

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SVP
Joined: 03 Feb 2003
Posts: 1619
Followers: 5

Kudos [?]: 27 [0], given: 0

PS: FREAK [#permalink]  19 Jun 2003, 02:01
00:00

Difficulty:

5% (low)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
15+Z=|Z-9|+Z!
Intern
Joined: 03 Feb 2003
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I want to say that there is no solution to this one.

15+Z=|Z-9|+Z!

if Z>=9 we will get 15+Z=Z-9+Z! or Z!=24 or Z=4 - not sefficient as Z>=9

if Z<9

15+Z=-Z+9+Z! , Z!-2Z=6 - no solution
Intern
Joined: 03 Feb 2003
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 0

checking...

15+Z=|Z-9|+Z!

Z=4

15+4=19
|4-9|+4!=5+24=29

19 =/= 29
SVP
Joined: 03 Feb 2003
Posts: 1619
Followers: 5

Kudos [?]: 27 [0], given: 0

you are right Z=4 is an extraneous root and should be terminated. No solution after all.
Intern
Joined: 15 Apr 2003
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FActorials Help!! [#permalink]  27 Jun 2003, 07:39
Z!-2Z=16

How can you solve this equation. Can someone show me the algebra or do you just have to plug in numbers for Z and use trial and error?

Also, when you see Z! do you automatically assume that Z is positive and an interger?

It seems like Z does not have to be an interger b/c you can have factorias of fractions such as 2.5! which would equal 2*1.

Can you have -4!
SVP
Joined: 03 Feb 2003
Posts: 1619
Followers: 5

Kudos [?]: 27 [0], given: 0

Yes, they are tricky; this is why they are here, at the forum.

Things are easy when you remember that a factorial is more or equal than ONE and that it is defined to NONNEGATIVE INTEGERS only. So, cheking a couple of integers would be enough. Likewise, you may draw two graphs to see intersections.

Z!=16+2Z
the tolerance range is {Z>=0
...........................and{16+2Z>=1; thus, Z>=0

then draw graphs inthe tolerance region and find intersection. Z=4, which is not a solution for the initial equation.
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