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Find an integer M. (1) M^M=M^3 (2) =1

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Manager
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 [#permalink] New post 26 May 2003, 00:06
bill gates trying to screw us on GMAT??? :)

i have to burst my bubble tho

0^0 =1, but 0^3=0, so x=0 is not a possible solution from (1)

originally i was thinking 0^0=0, and therefore a possible solution
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 [#permalink] New post 26 May 2003, 00:19
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I think Billy is right behind us


I tried multiplying 0 x 0 and such - as to the rules of powers (2^3 = 2x2x2) but it gives a 0. So far there are 3 choices for 0^0 (1, 0, and undefined, which I would claim makes it an undefined case, but we'll soon see. I am writing a short message to GMAC). It is obviously one of those made up things for the sake of sanity in the Universe.


-=-

Last edited by bb on 26 May 2003, 00:30, edited 1 time in total.
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 [#permalink] New post 26 May 2003, 00:27
here's my understanding:

0^x, where x>0, is always equal to 0

0^x, where x<0, is undefined

x^0, where x is ANY number (negative, positive, or zero) always equal to 1
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 [#permalink] New post 26 May 2003, 01:10
Guys, 0^0 ≠ 0^3.
0^0 has no solution, at least for the GMAT program.
Thus, negate it, eliminate it, terminate it, kill it, forget about it :yak
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COOL ! [#permalink] New post 01 Jun 2003, 22:41
The answer is E... I love this question.
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X, Y, and Z are integers. Find (X*Y)^Z. (1) 1=|X*Y| (2) [#permalink] New post 02 Jun 2003, 00:11
X, Y, and Z are integers. Find (X*Y)^Z.

(1) 1=|X*Y|
(2) Z^2=Z^3
:stupid2
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 [#permalink] New post 02 Jun 2003, 02:15
Who knows... Let us wait what other people say
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 [#permalink] New post 02 Jun 2003, 04:29
Ans : E

1) 1 or -1
2) 0 or 1

Combined : 1, 1, 1, or -1
  [#permalink] 02 Jun 2003, 04:29
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