Aptitude Discussion

Q. |
If $x$ and $y$ are any natural numbers, then which of the following is an odd number? |

✖ A. |
$x^y + y^x + (x-y)(x^y + x)$ |

✖ B. |
$x^y (x + y)(x^y + x)$ |

✖ C. |
$y^x (x^y – y) (x^y - x)$ |

✔ D. |
None of these |

**Solution:**

Option(**D**) is correct

$x$ and $y$ are natural numbers.

We know that of any natural number $p$.

$p^n + p$ is even

And, $p^n – p$ is even.

When we multiply an even number to any natural number the resultant number is even.

**Edit:** Based on comments, final answer has been changed from option A to option D.

**Pamish**

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**Vishal**

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take value $x=1$ and $y= 2$ as both are natural no. then it satisfy equation 1.

So answer should be option A.

No, you are wrong.

As Shireen explains in one of the comments here that this is one of the possible outcomes. But questions demands the number to be odd for ALL (or any) natural numbers.

Thus, it does not meet the requirement and hence not the correct answer.

**AJAY**

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take $x=1$ and $y=1$ and solve ....

we get odd numer by option A

SORRY, take, $X=1$ AND $Y=2$

Few of the values might give ODD numbers when putting values in option A, but this is NOT TRUE FOR ALL values.

And thus not a correct choice.

**Sanjay Singh**

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In option A) I agree that $(x^y+x)$ is even. Multiplying it with $(x-y)$ produces an even result since $(x^y+x)$ is even.

But what is the guarantee that $x^y$ or $y^x$ will always be even?

If any of these two is odd, then the entire result becomes odd, since,

even + odd = odd.

You are right Sanjay, but ODD number is what they have asked as a solution. Isn't it?

**Niket**

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I think, the answer should be none of these as in the first case the result would only be an odd number if at least x or y is an odd number otherwise the result will be an even number in the first case also.....

Please comment if anyone doesn't find it correct.

**Ankit**

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I think option will be "D".

I tried to put $x=2$ and $y=2$ in all of them but gives even number not odd number.

**Sumit**

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suppose $x=2$ , $y=1$

$x^y+y^x+(x-y)(x^y+x)$

$=2 +1 +1*3$

$=6$

Which is not odd.

Kindly Explain

1*3 how you have deducted this!!It is x^y+x and you have taken x=2 so 2+2=4 this it should be 1*4=4

and 4+1+2=7 which is odd

**Shreya**

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$pn =p$ $pn-p$ both will be even?

If you could explain ur question in a better way, I would be happy to help.

Yes shreya both $p^n + p$ and $p^n - p$ will be even .....

If we are getting EVEN outcome for option A even for 1 or 2 cases, it proves that it cannot satisfy being ODD for all cases. Since we have a few cases, where this gives us EVEN outcome.