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# which of the following gives the complete range of the

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Director
Joined: 26 Feb 2006
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which of the following gives the complete range of the [#permalink]  13 Jul 2007, 11:05
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which of the following gives the complete range of the values for which X! / Y! is an integer?

a) x > y
b) x > or = y
c) x < y
d) x < or = y
e) x = o
VP
Joined: 08 Jun 2005
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[#permalink]  14 Jul 2007, 02:18
Sorry to rock the boat , but what about:

X = 0
Y = 1

X < Y

0!/1! = 1

Senior Manager
Joined: 27 Jul 2006
Posts: 298
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[#permalink]  14 Jul 2007, 08:46
Killersquirell is right!!

Luckily there is a caveat in the GMAT. We have to pick the best answer provided, not always the only or right answer.
Director
Joined: 26 Feb 2006
Posts: 905
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[#permalink]  14 Jul 2007, 09:18
KillerSquirrel wrote:
Sorry to rock the boat , but what about:

X = 0
Y = 1

X < Y

0!/1! = 1

good point and perfectly make sense.

This is a TPR question. materials other than OG and MNHTN are not reliable. Therefore I am skeptical about TPR, Kaplan and other gmat materials..

thanx KillerSquirrel
Manager
Joined: 25 Jul 2006
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[#permalink]  14 Jul 2007, 14:47
nice catch killersquirrel!!!
just recently, i noticed a problem with a similar gotcha.
A+ for the answer
(extra credit) A++ for spotting the flaw in the answer...

If M and N are even ints, which of the following should also be an even int?

(M + 2)^|N|
(M*N)^|M - N|
N^|M| * N
M^N * M^2
(M + 1)*(N - 1)
Director
Joined: 26 Feb 2006
Posts: 905
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Kudos [?]: 52 [0], given: 0

[#permalink]  14 Jul 2007, 14:53
oops wrote:
nice catch killersquirrel!!!
just recently, i noticed a problem with a similar gotcha.
A+ for the answer
(extra credit) A++ for spotting the flaw in the answer...

If M and N are even ints, which of the following should also be an even int?

(M + 2)^|N|
(M*N)^|M - N|
N^|M| * N
M^N * M^2
(M + 1)*(N - 1)

C.

(M + 2)^|N| could be 1.
(M*N)^|M - N| could be 1.

N^|M| * N must be an even integer.

M^N * M^2 could be a fraction.
(M + 1)*(N - 1) is odd.

its a good question
Director
Joined: 18 Jul 2006
Posts: 531
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Kudos [?]: 30 [0], given: 0

[#permalink]  14 Jul 2007, 15:19
Himalayan wrote:
oops wrote:
nice catch killersquirrel!!!
just recently, i noticed a problem with a similar gotcha.
A+ for the answer
(extra credit) A++ for spotting the flaw in the answer...

If M and N are even ints, which of the following should also be an even int?

(M + 2)^|N|
(M*N)^|M - N|
N^|M| * N
M^N * M^2
(M + 1)*(N - 1)

C.

N^|M| * N must be an even integer.

For M, N = 0, answer should be 1, which is not even.
Manager
Joined: 25 Jul 2006
Posts: 97
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Kudos [?]: 0 [0], given: 0

[#permalink]  14 Jul 2007, 15:34
yes, good work juaz and himalayan!
the answer C, although correct for almost all cases, it fails when M and N are both 0, bcoz 0^0 is undefined and the result is NOT an even integer.

since u both found it together, i suggest you split the extra credit
Director
Joined: 18 Jul 2006
Posts: 531
Followers: 1

Kudos [?]: 30 [0], given: 0

[#permalink]  14 Jul 2007, 15:58
oops wrote:
yes, good work juaz and himalayan!
the answer C, although correct for almost all cases, it fails when M and N are both 0, bcoz 0^0 is undefined and the result is NOT an even integer.

since u both found it together, i suggest you split the extra credit

I agree that 0^0 could be any value (odd, even) , but I read smwhere that 0^0 = 1 is "widely accepted" now for practical purposes.
Try this on calculator and you'll get 1.

Ok...From Matforum:
The following is a list of reasons why 0^0 should be 1.
Rotando & Korn show that if f and g are real functions that vanish at the origin and are analytic at 0 (infinitely differentiable is not sufficient), then f(x)^g(x) approaches 1 as x approaches 0 from the right.

From Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik):
Some textbooks leave the quantity 0^0 undefined, because the functions 0^x and x^0 have different limiting values when x decreases to 0. But this is a mistake. We must define x^0=1 for all x , if the binomial theorem is to be valid when x=0 , y=0 , and/or x=-y . The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant.

Published by Addison-Wesley, 2nd printing Dec, 1988.
As a rule of thumb, one can say that 0^0 = 1 , but 0.0^(0.0) is undefined, meaning that when approaching from a different direction there is no clearly predetermined value to assign to 0.0^(0.0) ; but Kahan has argued that 0.0^(0.0) should be 1, because if f(x), g(x) --> 0 as x approaches some limit, and f(x) and g(x) are analytic functions, then f(x)^g(x) --> 1 .

The discussion of 0^0 is very old. Euler argues for 0^0 = 1 since a^0 = 1 for a not equal to 0 . Consensus has recently been built around setting the value of 0^0 = 1 .
Manager
Joined: 07 Jul 2007
Posts: 126
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Kudos [?]: 4 [0], given: 0

[#permalink]  14 Jul 2007, 22:02
0 is neither even or odd integer so it out of question. So answer is C
Director
Joined: 26 Feb 2006
Posts: 905
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Kudos [?]: 52 [0], given: 0

[#permalink]  14 Jul 2007, 22:57
N750 wrote:
0 is neither even or odd integer so it out of question. So answer is C

0 is always even.
logical reasoning:

even = even - 2
odd = odd - 2
odd - 2 =/= 0, therefore 0 is even.

properties of zero:
0! = 1
0^0 is not defined.
0 is neither positive nor negative
0 is even
X^0 = 1
0 is divisible by every integer (except 0), Since remainder of 0/k = 0
0 is a multiple of every integer. 0 = k*0

http://www.gmatclub.com/phpbb/viewtopic ... ht=#342549

for simplicity i have directed this question to here:
http://www.gmatclub.com/phpbb/viewtopic ... 460#343460
Intern
Joined: 13 Jul 2007
Posts: 8
Followers: 0

Kudos [?]: 3 [0], given: 0

[#permalink]  15 Jul 2007, 04:49
whether 0 is even is an old controversy, but we are not in abstract mathematics. OG 11 clearly includes 0 in its list of even numbers. thats all we need to know
[#permalink] 15 Jul 2007, 04:49
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# which of the following gives the complete range of the

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