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If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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24 Mar 2011, 06:54
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23% (01:58) correct 77% (01:51) wrong based on 302 sessions
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If m and k are nonzero integers and if y^(m+k) = y^m, what is the value of y? (1) k is odd. (2) y is odd.
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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27 Mar 2011, 03:31
rxs0005 wrote: If m and k are nonzero integers and
If y^(m+k) = y^m
what is the value of y?
(1) k is odd. (2) y is odd. The solutions above are not quite right. Before looking at the statements, If y^(m+k) = y^m, then we know that (y^m)(y^k) = y^m. There are now two possibilities: either y^k = 1, in which case y = 1 or y = 1, since k is nonzero, or y = 0. From Statement 1, we can be sure that y is not 1, but y can still be 0 or 1. From Statement 2, we can be sure that y is not 0, but y can still be 1 or 1 (if k is even). Using both Statements, we can be sure that y = 1 and the answer is C.
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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24 Mar 2011, 07:41
rxs0005 wrote: If m and k are nonzero integers and
If y^(m+k) = y^m
what is the value of y?
(1) k is odd. (2) y is odd. y^m*y^k = y^m y^k=1 (1) k=odd k can be 1,3,5,1. Means; y must be 1. Sufficient. (2)y=odd (1)^1=1 (1)^2=1 Not Sufficient. Ans: "A"
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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24 Mar 2011, 07:44
y^(m+k) = y^m => y^m x y^k = y^m => y^k = 1 Statement 1: If k is odd, then y = 1 Sufficient Statement 2: y is odd y can be +1 or 1 if the value of k is even and y can be +1 if the value of k is odd Not sufficient! Ans  A
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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27 Mar 2011, 03:52
IanStewart wrote: rxs0005 wrote: If m and k are nonzero integers and
If y^(m+k) = y^m
what is the value of y?
(1) k is odd. (2) y is odd. The solutions above are not quite right. Before looking at the statements, If y^(m+k) = y^m, then we know that (y^m)(y^k) = y^m. There are now two possibilities: either y^k = 1, in which case y = 1 or y = 1, since k is nonzero, or y = 0. From Statement 1, we can be sure that y is not 1, but y can still be 0 or 1. From Statement 2, we can be sure that y is not 0, but y can still be 1 or 1 (if k is even). Using both Statements, we can be sure that y = 1 and the answer is C. Thanks IanStewart. Wonder why rxs0005 kept mum!!!
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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27 Mar 2011, 16:41
Ian GMAT does not recognize 0^( anything) so with that in mind A should be the answer otherwise C would be correct and based on deductive reasons this kind of problem would not show up in GMAT where we need to exclude 0^
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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28 Mar 2011, 07:00
rxs0005 wrote: Ian GMAT does not recognize 0^( anything) so with that in mind A should be the answer otherwise C would be correct and based on deductive reasons this kind of problem would not show up in GMAT where we need to exclude 0^ The GMAT certainly does recognize 0^x, at least when x is a positive integer; that's always equal to zero. I'm not sure why you think that is undefined; 0^2, for example, is just equal to the product of two zeros, which is certainly zero. You may be thinking of the specific exception 0^0, which is not defined. In any case, the question probably should specify that m and k are positive integers to avoid exceptional cases.
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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28 Mar 2011, 09:24
It must be C. y^m*y^k = y^m y^m(y^k1)=0 so we know y=0/y=1/y=1 (when k is even) I  k is odd so y can still be 0 or 1  Insufficient. II  y is odd (so 1 or 1)  still insufficient. I+II  y is not 0  means y^k1=0  y^k=1 if K is odd  Y must be 1. so answer is C.
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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28 Mar 2011, 09:24
It must be C. y^m*y^k = y^m y^m(y^k1)=0 so we know y=0/y=1/y=1 (when k is even) I  k is odd so y can still be 0 or 1  Insufficient. II  y is odd (so 1 or 1)  still insufficient. I+II  y is not 0  means y^k1=0  y^k=1 if K is odd  Y must be 1. so answer is C.
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Number Properties
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01 Aug 2018, 11:10
Can someone please explain me this :
If m and k are nonzero integers and if y^(m+k) = y^m, what is the value of y?
(1) k is odd. (2) y is odd



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Re: Number Properties
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01 Aug 2018, 18:23
To find the value of y \(y^{m+k} = y^m\) => \(y^m * y^k = y^m\) => \(y^k = 1\) => k = 0 and y is any integer other than zero OR=> k = 1 and y is either 0 or 1 Statement 1 k is odd. If k is odd => k = 1 and y is either 0 or 1 No unique value for y Statement 1 is not sufficientStatement 2y is odd y can be 1 and k can be 1 satisfying \(y^k = 1\) y can be 5(or any integer other than 0) and k can be 0 satisfying \(y^k = 1\) No unique value for y Statement 2 is not sufficientCombining statements 1 and 2k is odd and y is odd => only k = 1 and y = 1 satisfy the given equations. Hence option C
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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01 Aug 2018, 19:26
workout wrote: To find the value of y
\(y^{m+k} = y^m\)
=> \(y^m * y^k = y^m\)
=> \(y^k = 1\)
=> k = 0 and y is any integer other than zero OR
=> k = 1 and y is either 0 or 1
Statement 1
k is odd.
If k is odd => k = 1 and y is either 0 or 1
No unique value for y
Statement 1 is not sufficient
Statement 2
y is odd
y can be 1 and k can be 1 satisfying \(y^k = 1\)
y can be 5(or any integer other than 0) and k can be 0 satisfying \(y^k = 1\)
No unique value for y
Statement 2 is not sufficient
Combining statements 1 and 2
k is odd and y is odd
=> only k = 1 and y = 1 satisfy the given equations.
Hence option C K cannot be taken as zero. In statement 2 y can take 2 values 1 or 1 Therefore its insufficient.
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If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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01 Aug 2018, 22:30
Koko11 wrote: Can someone please explain me this :
If m and k are nonzero integers and if y^(m+k) = y^m, what is the value of y?
(1) k is odd. (2) y is odd \(y^(m+k) = y^m\) \(y^m * y^k = y^m\) Therefore, \(y^k = 1\)  (1) From this we concur, For k we can have only 0 or 1 or 1, but as per premise, K is not 0 Hence k=1 or 1  (2) So if k = 1, From (1) we have y = 0 or 1  (3) and if k = 1, From (1) we have y = 1  (4) (y cannot be 0 as 0 raised to negative number will always be not defined) Now, Statement 1 : k is oddFrom (2) we already know that k is odd Hence Statement 1 Insufficient as there is no new informationStatement 2 : y is oddFrom (3), (4) and statement 2 we know, y = 1 Hence statement 2 alone is Sufficient.IMO it should be B.
But OA says C.
Does not seem to be correct
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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01 Aug 2018, 22:51
adstudy wrote: Koko11 wrote: Can someone please explain me this :
If m and k are nonzero integers and if y^(m+k) = y^m, what is the value of y?
(1) k is odd. (2) y is odd \(y^(m+k) = y^m\) \(y^m * y^k = y^m\) Therefore, \(y^k = 1\)  (1) From this we concur, For k we can have only 0 or 1 or 1, but as per premise, K is not 0 Hence k=1 or 1  (2) So if k = 1, From (1) we have y = 0 or 1  (3) and if k = 1, From (1) we have y = 1  (4) (y cannot be 0 as 0 raised to negative number will always be not defined) Now, Statement 1 : k is oddFrom (2) we already know that k is odd Hence Statement 1 Insufficient as there is no new informationStatement 2 : y is oddFrom (3), (4) and statement 2 we know, y = 1 Hence statement 2 alone is Sufficient.IMO it should be B.
But OA says C.
Does not seem to be correctIn statement 2 Y could be 1, and therefore it is insufficient
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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02 Aug 2018, 04:14
Raksat wrote: adstudy wrote: Koko11 wrote: Can someone please explain me this :
If m and k are nonzero integers and if y^(m+k) = y^m, what is the value of y?
(1) k is odd. (2) y is odd \(y^(m+k) = y^m\) \(y^m * y^k = y^m\) Therefore, \(y^k = 1\)  (1) From this we concur, For k we can have only 0 or 1 or 1, but as per premise, K is not 0 Hence k=1 or 1  (2) So if k = 1, From (1) we have y = 0 or 1  (3) and if k = 1, From (1) we have y = 1  (4) (y cannot be 0 as 0 raised to negative number will always be not defined) Now, Statement 1 : k is oddFrom (2) we already know that k is odd Hence Statement 1 Insufficient as there is no new informationStatement 2 : y is oddFrom (3), (4) and statement 2 we know, y = 1 Hence statement 2 alone is Sufficient.IMO it should be B.
But OA says C.
Does not seem to be correctIn statement 2 Y could be 1, and therefore it is insufficient With Y as 1 we can get \(y^k = 0\) only if k = 0 and k cannot be zero as stated in the premise. Hence i considered it to be sufficient
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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02 Aug 2018, 07:08
adstudy why can't K be 2? Then y could be 1. Therefore we need statement 1 so that we could reach 1 unique solution. That is y=1 Therefore C is correct
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Re: If m and k are nonzero integers and if y^(m+k) = y^m, what is the
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02 Aug 2018, 23:50
Raksat wrote: adstudy why can't K be 2? Then y could be 1. Therefore we need statement 1 so that we could reach 1 unique solution. That is y=1 Therefore C is correct thanks Raksat for pointing this out. I missed on that. Yes the answer now seems to be C
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