If x and y are positive integers, is x even?

(1) for the sum to be even....both terms should be even/odd......for that...both X,y should be either even or odd..... insufficient

(2) however 4x will be even whatever may the value of X....so y^x should be odd for the sum to be odd....so for which X,y has to take (odd/even,even/odd) or (odd,odd).....again not helpful..... insufficient


Combining we get we knew that one term is odd from (2) so first term should also be odd for which X should be odd


OA:C

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(1) shows that x and y must be either both odd or both even. We don't know which one.
NOT SUFFICIENT.

(2) shows that y is odd. We don't have any useful information on x
NOT SUFFICIENT.

(1)+(2)
We know that y is odd from (2) and, as a result, both x and y must be odd from (1). Thus, x is odd
SUFFICIENT

FINAL ANSWER IS (C)

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#1
x^y+y^x is even
possible when either x ,y are both odd or even ; insufficeint
#2
y^x+4x is odd.
y has to be odd and x can be either odd or even
from 1 &2
if y is odd then x has to be odd as well
IMO C


if x and y are positive integers, is x even?

(1) x^y+y^x is even.

(2) y^x+4x is odd.

If x and y are positive integers, is x even?

(1) \(x^y+y^x\) is even.
\(o^o + o^o = e\)
\(e^e + e^e = e\)

INSUFFICIENT.

(2) \(y^x+4x\) is odd.
\(o^o + e.o = o\)
\(o^e + e.e = o\)

INSUFFICIENT.

Together 1 and 2 only case left is
\(o^o + e.o = o\) for \(y^x+4x\)
\(o^o + o^o = e\) for \(x^y+y^x\)
x is odd.

SUFFICIENT.

Answer C.

If x and y are positive integers, is xx even?

(1) \(x^y+y^x\) is even.
=> both x and y are odd or both x and y are even
=> Not suff


(2) \(y^x+4x\) is odd.
=> \(y^x\) is odd
=> y is odd whether x is odd or even.
=> Not suff

Combine (1) and (2) => y is odd and x is odd too.
=> Suff
=> Choice C

If x and y are positive integers, is x even?

(1) x^y + y^x is even.

(2) y^x+4x is odd.
Solution:
From statement (1), x^y + y^x is even, The value of x and y together as base both can be even or odd . In this case, the value of y and x as exponential power has no value. So we cannot definitely say whether x is even or not. Statement is insufficient.

From statement (2), y^x+4x is odd. Here the value of y^x must be odd. Hence y is odd.
But x may be even or odd. Not sufficient.

Combining statement (1) and (2), As y is odd, x must be odd.
Answer C)

(1) insufic
x,y=(odd,odd)(even,even)

(2) insufic
x=any
y=odd

(1,2) sufic
y=odd:
odd^x + x^odd = even,
odd+odd = even,
x=odd

Ans (C)

If x and y are positive integers, is xx even?

(1) x^y+y^x is even.

(2) y^x+4x is odd.

1) The sum of two integers can be even when both of them are either odd or even. In this case if x is even, y can be even or odd, or if x is odd, y has to be odd. Since more than one possibilities are open, this is not sufficient.

2) For the sum of two integers to be odd, one of them has to be odd and another has to even. Here, 4x will always be even, so y^x has to be odd. Only odd integers with odd multitudes can be an odd integer. so both x and y has to be odd. sufficient.

B is the answer.

We are given that x and y are positive integers. We are to determine if x is even.

Statement 1: x^y + y^x is even.
There are two conditions for which x^y + y^x results in an even number.
i. x^y is odd and y^x is odd so that odd+odd=even
or ii. x^y is even and y^x is even so that even+even=even
In i. above, x and y must necessarily be odd. This is because an odd number multiplied by itself any positive integral number of times will yield an odd number. so if x is odd, and y is even or odd, x^y will be odd. Similarly, if x is even and y is odd or even, x^y will always be even.
So either both x and y are odd or both x and y are even, leading to yes and no answers.
Statement 1 is insufficient.

Statement 2: y^x + 4x is odd.
from statement 2, we can clearly say that y is odd. But x can be either odd or even and 4x will always be even.
Statement 2 is insufficient.

1+2
Sufficient.
We know that y is odd, so x must also be odd per the caveat that statement 1 has two possibilities. Both x and y are either even or odd. So if y is odd, x is odd. We know y is odd from statement 2, hence x must be odd as well.

The answer is C.

If x and y are positive integers, is x even?

(Statement1): \(x^{y} + y^{x}\) is even.
If x=1,y=1, then \(1^{1}+ 1^{1}= 2\) (x—odd)
If x=2, y=2, then \(2^{2}+2^{2} = 8\) (x—even)
Insufficient

(Statement2): \(y^{x} + 4x\) is odd.
In order \(y^{x} +4x\) to be odd number, y must be odd integer.
— x could be odd or even integer
Insufficient

Taken together 1&2,
y is odd integer —> in order \(x^{y}+y^{x}\) to be even number, x must be always odd number.
—> Always NO
Sufficient

The answer is C

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Explanation:

Given: x and y are positive integers

Statement 1:x^y+y^x is even.
So take values of (x,y) = (1,1) ; (2,2)
The above values will give result as even. but x can be odd or even.

Not sufficient.

Statement 2: y^x+4x is odd.
So, 4x part will be even always, x can be anything.
Y has to be odd for the resultant. So, x can be 1,2,3 anything....
Not sufficient.

Combining 1&2;

As y is odd, x must be odd.

IMO-C

If \(x\) and \(y\) are positive integers, is \(x\) even?

(1) \(x^y + y^x\) is even.
x & y are both either even or odd.
Since it is unknown whether y is even or odd
NOT SUFFICIENT

(2) \(y^x + 4x\) is odd.
since 4x is even; y^x is odd -> y is odd
Still x may be even or odd.
NOT SUFFICIENT

(1) + (2)
(1) \(x^y + y^x\) is even.
x & y are both either even or odd.
(2) \(y^x + 4x\) is odd.
since 4x is even; y^x is odd -> y is odd
x is also odd since \(x^y + y^x\) is even.
SUFFICIENT

IMO C