If x and y are positive integers and x^y = y^x + 1, which of the follo

Question

If x and y are positive integers and  x^y = y^x + 1 , which of the following must be odd?

A. xy
B. x + y - 1
C. x + 2y
D. 3x - y
E. xy + 2

M37-53

Bunuel · Accepted Answer

Official Solution:

If  x  and  y  are positive integers and  x^y = y^x + 1 , which of the following must be odd?

A.  xy 
B.  x+y-1 
C.  x+2y 
D.  3x-y 
E.  xy+2

CASE 1: if  x=odd , then  x^y = odd  and thus  y^x   must be even, so  y  must be even.

CASE 2: if  x=even , then  x^y = even  and thus  y^x   must be odd, so  y  must be odd.

A.  xy  is even in both cases.

B.  x+y-1  is even in both cases.

C.  x+2y  is odd in the first case and even in the second case.

D.  3x-y  is odd in both cases.

E.  xy+2  is even in both cases.

Answer: D

Raxit85 · Answer

Let me try with no. Plugging method which can satisfy the equation, X=3, y=2.
So, by hit and trial, option C gives odd no. Ans. C

Posted from my mobile device

RupamPaul13 · Answer

IMO D. For the equation to follow, there can be 2 condition.

1. X odd and Y even
2. X even and Y odd.

By substitution of the above we can get that option D 3x-y will always be odd for any of the above two scenario. Hence D is the correct choice.

Posted from my mobile device

Vinit800HBS · Answer

Raxit85

I am afraid that you didn’t check all the options. In fact, for the set of values for X and Y given by you, Option D also yields odd number.

Coming to the question,

There are two cases that we have to consider that ensure the relationship is intact.

Case 1: X = 3 and Y = 2
For this case, Option C and D both yield odd number.

Case 2: X = 2 and Y = 1
For this case, only Option D gives us odd number.

So, the answer has to be Option D.

Hope that helps..

Cheers ?

KSBGC · Answer

x^y = y^x + 1

This equation holds true only if x = 2 and y = 1.

Note:

both x and y are positive integers.

x^y = y^x + 1

x^y - y^x = 1

observe that difference is odd. It clearly indicates that x and y must be different.

Minimum value of x and y is 1. A bit number sense is required.

Option D is the correct answer.

Archit3110 · Answer

let x=2 , y=1
 x^y = y^x + 1 
2^1=1^2+1
2=2

for value 3x-y ; 6-1 ; 5
IMO D

nick1816 · Answer

x^y=y^x+1
Case 1- if y is even then x is odd
Case 2. If y is odd then x is even

1. xy = E*O=E or O*E=E
2. x+y-1=O+E-1=E or E+O-1=E
3.x+2y= O+2*E=O or E+2*O=E
4. 3x-y= 3*O-E=O or 3*E-O=O
5. xy+2= O*E+2=E or E*O+2=E

Only option D is odd in both cases

exc4libur · Answer

O*O=O;
E*(O,E)=E;
O+E=O;
(E+-E,O+-O)=E

y=O:x^y = y^x + 1…x = O + O=E…(x,y)=(E,O) 
 y=E:x^y = y^x + 1…x = E + O=O…(x,y)=(O,E)

A. xy: OE=E
B. x + y - 1: O+E-O=O-O=E
C. x + 2y: O+2(E)=O+E=O; E+2(O)=E+E=E
E. xy + 2: OE+E=E; EO+E=E

Ans (D):  3(O)+E=O+E=O; 3(E)+O=E+O=O

Rocknrolla21 · Answer

Hi, Experts!

In such questions, how many values should we check?

I just checked x=2 and y=1. These values worked, and I got the right answer. If we get only one answer using a particular set of numbers, should we test more values? I make a lot of mistakes in such questions that require you to test values.

Thanks!

sujoykrdatta · Answer

Since it is an odd-even question, we will focus on whether x or y are even / odd:

x^y = y^x + 1

Case 1: If x is odd:  
LHS is odd raised to some positive integer power = odd
Thus, RHS must be odd =>  y^x  must be even => y is even

Checking the options: 
A. xy = Odd * Even = Even
B. x + y - 1 = Odd + Even - 1 = Even
C. x + 2y = Odd + Even = Odd
D. 3x - y = Odd - Even = Odd
E. xy + 2 = Odd * Even + 2 = Even

Thus, the answer must be either Option C or Option D

Case 2: If x is even:  
LHS is even raised to some positive integer power = even
Thus, RHS must be even =>  y^x  must be odd => y is odd

Checking the options C and D only: 
C. x + 2y = Even + Even = Even
D. 3x - y = Even - Odd = Odd

Thus, the answer must be Option D

Answer D

Adarsh_24 · Answer

case1 and case2 should end in conclusions about y. May be a typo?

Bunuel · Answer

Absolutely! Thank you for oticing. Edited.

DanTheGMATMan · Answer

https://www.youtube.com/watch?v=Yl1i9txiVF0

If x and y are positive integers and x^y = y^x + 1, which of the follo

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