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If m and k are non-zero integers and if y^(m+k) = y^m, what is the

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If m and k are non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 24 Mar 2011, 07:54
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If m and k are non-zero 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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 27 Mar 2011, 04:31
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rxs0005 wrote:
If m and k are non-zero 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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 24 Mar 2011, 08:41
1
rxs0005 wrote:
If m and k are non-zero 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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 24 Mar 2011, 08:44
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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 27 Mar 2011, 04:52
IanStewart wrote:
rxs0005 wrote:
If m and k are non-zero 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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 27 Mar 2011, 05:13
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Re: If m and k are non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 27 Mar 2011, 17:41
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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 28 Mar 2011, 08:00
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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 28 Mar 2011, 10:24
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It must be C.

y^m*y^k = y^m

y^m(y^k-1)=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^k-1=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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 28 Mar 2011, 10:24
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It must be C.

y^m*y^k = y^m

y^m(y^k-1)=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^k-1=0 ----- y^k=1
if K is odd - Y must be 1.

so answer is C.
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Number Properties  [#permalink]

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New post 01 Aug 2018, 12:10
Can someone please explain me this :

If m and k are non-zero 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  [#permalink]

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New post 01 Aug 2018, 19:23
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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
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Re: If m and k are non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 01 Aug 2018, 20:26
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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 01 Aug 2018, 23:30
Koko11 wrote:
Can someone please explain me this :

If m and k are non-zero 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 odd

From (2) we already know that k is odd

Hence Statement 1 Insufficient as there is no new information

Statement 2 : y is odd

From (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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 01 Aug 2018, 23:51
adstudy wrote:
Koko11 wrote:
Can someone please explain me this :

If m and k are non-zero 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 odd

From (2) we already know that k is odd

Hence Statement 1 Insufficient as there is no new information

Statement 2 : y is odd

From (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

In statement 2 Y could be -1, and therefore it is insufficient
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Re: If m and k are non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 02 Aug 2018, 05:14
Raksat wrote:
adstudy wrote:
Koko11 wrote:
Can someone please explain me this :

If m and k are non-zero 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 odd

From (2) we already know that k is odd

Hence Statement 1 Insufficient as there is no new information

Statement 2 : y is odd

From (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

In 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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 02 Aug 2018, 08:08
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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 non-zero integers and if y^(m+k) = y^m, what is the  [#permalink]

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New post 03 Aug 2018, 00:50
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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Re: If m and k are non-zero integers and if y^(m+k) = y^m, what is the &nbs [#permalink] 03 Aug 2018, 00:50
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