|
Author |
Message |
|
TAGS:
|
|
|
SVP
Joined: 05 Jul 2006
Posts: 1563
Followers: 4
Kudos [?]:
63
[0], given: 34
|
if x and y are intigers is x^y * y^-x = 1?? 1) x^x > y 2) [#permalink]
 16 Sep 2006, 04:59
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
if x and y are intigers is x^y * y^-x = 1??
1) x^x > y
2) x>y^y
|
|
|
|
|
|
|
Director
Joined: 13 Nov 2003
Posts: 811
Location: BULGARIA
Followers: 1
Kudos [?]:
6
[0], given: 0
|
I used following
A) X^X=Y then stem is =1 but since it is inequality, then it is not = to 1
IMO it should be D
|
|
|
|
|
|
SVP
Joined: 01 May 2006
Posts: 1837
Followers: 8
Kudos [?]:
71
[0], given: 0
|
For me, it's (B)
Statment 1:
If x = y = 2, then :
> 2^2 = 4 > 2
> 2^2 / (2^2) = 1
If x = 2 and y = 1, then :
> 2^2 = 4 > 1
> 2^1 / (1^2) = 2
INSUFF
Statment 2:
x > y^y thus x > y (x & y are integers) that means x != y
Hence, x^y * y^-x != 1
SUFF
|
|
|
|
|
|
VP
Joined: 28 Mar 2006
Posts: 1396
Followers: 1
Kudos [?]:
14
[0], given: 0
|
Re: Excellent iniquality DS [#permalink]
16 Sep 2006, 07:23
yezz wrote: if x and y are intigers is x^y * y^-x = 1??
1) x^x > y
2) x>y^y
Can some one see if my working is correct?
given x,y are ints
is x^y * y^-x = 1
or x^y / y^x = 1 ===> x^y = y^x
taking logs on both sides we have to prove (y/x)(logx/logy)=1
from (1) we have
xlogx > log y
logx/logy > 1/x
(so we do not know anything about it)
from (2) we have
x>y^y ==> logx > ylogy
Means
logx/logy > y
Then I am lost... 
|
|
|
|
|
|
Director
Joined: 13 Nov 2003
Posts: 811
Location: BULGARIA
Followers: 1
Kudos [?]:
6
[0], given: 0
|
Hallo trivikram
think that ; or x^y / y^x = 1 ===> x^y = y^x
is NOT correct .It should be (x^y-y^x)/y^x=0 where y^x can not be 0 since division by 0 is not defined
|
|
|
|
|
|
VP
Joined: 28 Mar 2006
Posts: 1396
Followers: 1
Kudos [?]:
14
[0], given: 0
|
BG wrote: Hallo trivikram
think that ; or x^y / y^x = 1 ===> x^y = y^x
is NOT correct .It should be (x^y-y^x)/y^x=0 where y^x can not be 0 since division by 0 is not defined
But BG from there how do we proceed?
|
|
|
|
|
|
VP
Joined: 21 Aug 2006
Posts: 1026
Followers: 1
Kudos [?]:
12
[0], given: 0
|
Re: Excellent iniquality DS [#permalink]
16 Sep 2006, 08:37
yezz wrote: if x and y are intigers is x^y * y^-x = 1??
1) x^x > y
2) x>y^y
The condition x^y * y^-x = 1 will be valid only if x and y are 2 and 4
from statement 2 we can say that, that is not the case. Hence B is the anwer.
_________________
The path is long, but self-surrender makes it short;
the way is difficult, but perfect trust makes it easy.
|
|
|
|
|
|
Senior Manager
Joined: 31 May 2006
Posts: 379
Location: Phoenix AZ
Followers: 1
Kudos [?]:
4
[0], given: 0
|
......................... 
|
|
|
|
|
|
SVP
Joined: 01 May 2006
Posts: 1837
Followers: 8
Kudos [?]:
71
[0], given: 0
|
mailtheguru wrote: ......................... What's going on ? 
|
|
|
|
|
|
Current Student
Joined: 28 Dec 2004
Posts: 3439
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 11
Kudos [?]:
134
[0], given: 2
|
for me B is sufficient...
X and Y are integers... we dont know if they are postive or negative...
x^y * y^-x=1?
(1) x^x>y
well say if x and y=2...then 2^2>2...
2^2*1/(2^2)=1; x=2, y=3 then the stem isnt equal to 1
(2)
x>y^y; so we know that |x|>|y| and they are not equal..therefore the stem will never be 1....
sufficient
|
|
|
|
|
|
Senior Manager
Joined: 31 May 2006
Posts: 379
Location: Phoenix AZ
Followers: 1
Kudos [?]:
4
[0], given: 0
|
Fig wrote: mailtheguru wrote: ......................... What's going on ? Got screwed 
|
|
|
|
|
|
SVP
Joined: 01 May 2006
Posts: 1837
Followers: 8
Kudos [?]:
71
[0], given: 0
|
mailtheguru wrote: Fig wrote: mailtheguru wrote: ......................... What's going on ? Got screwed It's one part of the joy of Maths (or DS?  ).... One time u another time someone else (this rule works with me as well )
|
|
|
|
|
|
Senior Manager
Joined: 31 May 2006
Posts: 379
Location: Phoenix AZ
Followers: 1
Kudos [?]:
4
[0], given: 0
|
Fig wrote: mailtheguru wrote: Fig wrote: mailtheguru wrote: ......................... What's going on ? Got screwed It's one part of the joy of Maths (or DS? ).... One time u another time someone else (this rule works with me as well ) yeah, sure.
Where are these questiosn from?
|
|
|
|
|
|
Director
Joined: 06 May 2006
Posts: 791
Followers: 3
Kudos [?]:
7
[0], given: 0
|
A very shaky B for me as well...
x^y = y^x
=> ylogx = x logy
=> y/x = logy/logx -- (1)
A# x^x > y
=> xlogx > logy
=> From (1), y/x^2 = logy/xlogx < 1
=> y < x^2
Now, if x=2, we have y = 1, 2, 3... if y = 2 then x^y = y^x, but not in the case of y =1 or 3. Hence not sufficient.
B# x > y^y
=> logx > ylogy
=> from (1), y^2/x = ylogy/logx < 1
=> y^2 < x
Now, if x = 27, we have y = 1, 2, 3, 4 or 5; but the expression in the question does not evaluate to 1 in any of the cases. Hence sufficient.
|
|
|
|
|
|
SVP
Joined: 05 Jul 2006
Posts: 1563
Followers: 4
Kudos [?]:
63
[0], given: 34
|
if x and y are intigers is x^y * y^-x = 1??
1) x^x > y
2) x>y^y
rephrase stem x^y *1/y^x = 1 ie: receprocals
This is only possible if x^y = y^x and in turn this is only possible if x=y or x,y belongs to the set {2,4}
from one
x^x > y thus sure x,y are not equall one
x,y could be anything and could be x=4 and y=2 .....insuff
from two
x>y^ythus sure x,y are not equall one and sure /x/> /y/ thus if they are 2,4 x must be the 4 and thus y^y = 4....suff
because now we are sure that the only two conditions that make the assumption true are not valid ( being one or {2,4})
suff
answer is not a shaky b it is B 
Last edited by yezz on 16 Sep 2006, 13:35, edited 2 times in total.
|
|
|
|
|
|
SVP
Joined: 01 May 2006
Posts: 1837
Followers: 8
Kudos [?]:
71
[0], given: 0
|
yezz wrote: if x and y are intigers is x^y * y^-x = 1?? 1) x^x > y 2) x>y^y rephrase stem x^y *1/y^x = 1 ie: receprocals This is only possible if x^y = y^x and in turn this is only possible if x=y=1
or x,y belongs to the set {2,4}from one x^x > y thus sure x,y are not equall one x,y could be anything and could be x=4 and y=2 .....insuff from two x>y^ythus sure x,y are not equall one and sure /x/> /y/ thus if they are 2,4 x must be the 4 and thus y^y = 4....suff because now we are sure that the only two conditions that make the assumption true are not valid ( being one or {2,4}) suff answer is not a shaky b it is B Not agree with the bold
x = y makes it and no restricted values on it apart 0^0 (recently identified as not fitted for GMAT )
|
|
|
|
|
|
SVP
Joined: 05 Jul 2006
Posts: 1563
Followers: 4
Kudos [?]:
63
[0], given: 34
|
come on Fig you ve got to get over this exponent DS ...
and i agree with you on your comment but still it would be insuff
I will edit it
|
|
|
|
|
|
SVP
Joined: 01 May 2006
Posts: 1837
Followers: 8
Kudos [?]:
71
[0], given: 0
|
It's the end of my day here I'm allowed to turn the end of my post to something funny for those who know it
|
|
|
|
|
|
SVP
Joined: 05 Jul 2006
Posts: 1563
Followers: 4
Kudos [?]:
63
[0], given: 34
|
Man GMAT is funny
funny wn we do silly mistakes " kinder garten ones" apart from the test
and funny when you laugh yourself to death infront of a hard DS that you ve no clue about
we need those jokes all the time
by the way i like your audi
|
|
|
|
|
|
SVP
Joined: 01 May 2006
Posts: 1837
Followers: 8
Kudos [?]:
71
[0], given: 0
|
Ahhh If u start to speak about Audi... I'm not ready to make an end to this day finally
Frankly, I recommand this brand over the 2 other major german brands But I try to convince no one
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
