(1) States \(K^{x+y}=1\). If \(K = 1, x + y\) can be odd or even. Not sufficient.

(2) Nothing about \(x + y\). Not sufficient.

(1) and (2) together: Sufficient, because necessarily \(x + y = 0\).

Answer C.
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x+y is not necessarily zero. If k=-1, then x+y can be ANY even number.
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You are absolutely right. Nothing stated about K! At least the answer is still C.
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Thanks for the answer.

FROM STEM , X IS +VE but x+y doesnt have to be +ve , is x+y even??

from 1

k = 1 and x+y = anything or k= -1 , x+y = even or k is anything and x+y = 0 ... insuff

from 2

k is not 1 ... insuff

both together

k can = 1 and (x+y = even ) or k = anything and (x+y = 0) .... in either case x+y is even ... suff

C

Okay
so we need to get if x+y is even or not
x=> integer
remember => y is not specified as to being an integer.
Lets ook at statements

Statement 1
k^x*k^y=> k^(x+y)=> 1
okay if k is one then x+y can be anything => even/odd/fraction etc
hence not sufficient.
Statement 2
k≠1
no clue of x and y hence not sufficient
Lets Combine the statements
Hmm
K≠1
if k=-1 => x+y must be even as (-1)^even =1 and (-1)odd = -1
if k is any other integer except 1 and -1 then x+y must be zero.
But hey, zero is an even too.
BINGO
x+y must be always either 0 or even
Hence C
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Statement 1: K^x*K^y =1

K^(x+y) = 1, therefore

K = 1, then x+y can odd/even
K > 1, then x+y = 0, hence even
K = -1, then x+y is even
K < -1, then x+y = 0 hence even

Statement 1 is not sufficient.



Statement 2: K is not equal to 1.

No information about x or y. Statement 2 is not sufficient.


Combining, we get K = -1, > 1 or < -1, hence x+y is even

Combining is sufficient.


Answer C.


Thanks,
GyM
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