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8212;how about this one? |Z|+|Z1|=Z

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8212;how about this one? |Z|+|Z1|=Z [#permalink] New post 20 Jun 2003, 07:03
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—how about this one?

|Z|+|Z–1|=Z!
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 [#permalink] New post 20 Jun 2003, 11:15
{0, 1} are OK now. Post your solution, not wild guessing.
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 [#permalink] New post 12 Jun 2004, 00:15
A year is soon to pass by, but I see no solution here...
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Re: PS: MODUL+FACTORIAL-2 [#permalink] New post 12 Jun 2004, 00:27
stolyar wrote:
—how about this one?

|Z|+|Z–1|=Z!


Z is natural number (or 0) => if Z >= 3 => Z! >= 6 and grows much faster than 2*Z - 1. => for Z >= 3, Z! > 2*Z - 1.

Z belongs to the set {0, 1, or 2}.

Z = 0: OK.

Z = 1: OK.

Z = 2: NO...

=> Z = {0, 1}.
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 [#permalink] New post 13 Jun 2004, 12:34
yes, since Z is factorialized, it has to be a whole, nonnegative number. We can safely open |Z|.

Z+|Z-1|=Z!

Open the second modul.

1) if Z>1, then Z+Z-1=Z! or 2Z-1=Z!; no solution for whole Z>1
why? if Z>1, then the left part is always odd, while the right part is always even.

2) if 0<=Z<=1, then Z-Z+1=Z! or Z!=1; Z=[0 or 1]
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 [#permalink] New post 17 Jun 2004, 08:55
lastochka wrote:
I don't see 1) or 2)
weird...


lastochka, stolyar merely stated that z-1 is greater than 0 or lower than 0.

The first case is equivalent to |z-1| = z - 1, the second one is equivalent to |z-1| = 1-z. Then he considers these 2 cases.
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 [#permalink] New post 18 Jun 2004, 07:01
lastochka wrote:
I don't see 1) or 2)
weird...


Yes, many of my questions are weird and eerie. I like short but nonstandard problems. If you look back at the start of this forum, you will find tens of such questions.
  [#permalink] 18 Jun 2004, 07:01
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8212;how about this one? |Z|+|Z1|=Z

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