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# try this one—find all its roots. be careful not to

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SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

try this oneâ€”find all its roots. be careful not to [#permalink]  20 Jun 2003, 07:02
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try this one—find all its roots. be careful not to lose something.

|C|–|C–1|=C!
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

[#permalink]  20 Jun 2003, 10:58
Your 0 is FOREVER OUT! Try 0 and you will see that C!=–1, which is a nonsense! ANY factorial is always more than or equal to ONE.

You, guys, do not even give yourselves a job to check each root—but doing so is a REAL MUST while dealing with moduls. Moreover, you continue in the same way after my many long notations! Shame on you. I will ask Bogdan to downgrade your titles.

Post your approaches here. Let us locate your weak skills.
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 1

Kudos [?]: 1 [0], given: 0

PS: MODUL+FACTORIAL [#permalink]  26 Jun 2003, 05:24
The only root is 1

That is the only place where the two graphs intersect.
SVP
Joined: 03 Feb 2003
Posts: 1608
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Kudos [?]: 76 [0], given: 0

[#permalink]  26 Jun 2003, 21:50
Another way is classical - find mod zeroes, intersect lines, and open moduls:

____________0___________________1________________

–C–1+C=C!..........C–1+C=C!.....................C–C+1=C!
C1=–1..................2C–1=C! .......................C1=1
no solution............factorials are for ............C=0 – not in the interval
............................integers of which............C=1 OK
............................there are 0 and 1
............................0 is out, 1 is OK

Finally, after checking, we see that C=1.
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 1

Kudos [?]: 1 [0], given: 0

PS: MODUL+FACTORIAL [#permalink]  26 Jun 2003, 22:20
Stolyar, I like your approach better.

Thanks!
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

[#permalink]  26 Jun 2003, 22:40
sure, but sometimes long
I use it every time!
[#permalink] 26 Jun 2003, 22:40
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# try this one—find all its roots. be careful not to

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