If a and b are distinct non zero numbers, is x+y an even

Question

If a and b are distinct non zero numbers, is x+y an even integer ?

(1)  a^x*a^y = 1 .
(2)  b^x*b^y = 1 .

Bunuel · Accepted Answer

Note that: Any nonzero number to the power of 0 is 1:  a^0=1 , and one raised to any power is one:  1^n=1 .

(1) a^x*a^y = 1 -->  a^{x+y}=1  --> either  x+y=0=even  and in this case  a  can be any non zero number  or   a=1  and in this case  x+y  can equal to any number even odd or non-integer. Not sufficient.

(2) b^x*b^y = 1 --> the same here:  b^{x+y}=1  --> either  x+y=0=even  and in this case  b  can be any non zero number  or   b=1  and in this case  x+y  can equal to any number even odd or non-integer. Not sufficient.

(1)+(2) As  a  and  b  are  distinct  numbers then both cannot be equal to 1, so  x+y=0=even . Sufficient.

Answer: C.

eshan429 · Answer

You forgot to include one possibility when either a or b is -1. In this case also  x+y=even  should be true for  a^x*a^y = 1  or  b^x*b^y = 1  to hold but just wanted to mention that this is another possibility as the question does not restrict a & b to be positive.

ssriva2 · Answer

Hey  Bunuel ,
I know what im going to ask will sound a bit silly!
but if we combine both statements,we can also conclude that a=b=1 .Therefore, when a=b=1 and x+y=3= ans is 1
or a=b=1 and x+y=2=ans is 1

EgmatQuantExpert · Answer

Hi  ssriva2 ,

It is given in the question statement that a and b are distinct numbers. Hence a = b is not a valid conclusion.

Regards
Harsh

EgmatQuantExpert · Answer

The question tests the conceptual understanding of the cases where  base^{exponent} = 1 . A number raised to a power( say  p^q)  is equal to 1 in the following cases:

Case-I:  p = 1  and  q  is any number  
1 raised to any power will be equal to 1.

Case-II:  p = -1  and  q  is any even number  
-1 raised to an even power will be equal to 1

Case-III:  p  is any non-zero number and  q = 0   
Any non-zero number raised to power zero will be equal to 1

With this understanding let's analyze the statements.

Analyze Statement-I independently  
St-I tells us that  a^{x+y} = 1 . So, there are three possible cases:

i.  a = 1  and  x+y  is either even/odd
ii.  a = -1  and  x+y  is even
iii.  a  is any number and  x+y =0  which is even

Thus, statement-I does note tell us for sure if  x+y  is even or odd

Analyze Statement-II independently  
St-II tells us that  b^{x+y} = 1 . So, there are three possible cases:

i.  b = 1  and  x+y  is either even/odd
ii.  b = -1  and  x+y  is even
iii.  b  is any number and  x+y =0  which is even

Thus, statement-II does note tell us for sure if  x+y  is even or odd

Combining statements-I & II  
Combining st-I & II gives us two scenarios:

i.  a,b  are any numbers and  x+y = 0  which is an even
ii.  a = 1  and  b = -1  or vice versa. For both cases  x+y  will have to be an even for  -1^{x+y} =1 . Since a, b are distinct numbers we can say that  a  and  b  both can't be equal to 1 at the same time.

Both the scenarios tell us that  x+y  will be an even. Thus, combination of both the statements is sufficient to give us an unique answer.

Answer: Option C

Takeaway   
  p^q = 1  is one of the special cases in algebra. It's important to consider all its possible scenarios to arrive at the right answer.

Regards
Harsh

ssriva2 · Answer

Ohhh yes

. Completely missed this info while solving the question! :oops:

KrishnakumarKA1 · Answer

Prompt analysis
A,b, x and y are real numbers

Superset
The answer will either yes or no

Translation
In order to find the answer, we need:
1# exact value of x and y
2# any equation to get the characteristics of x and y

Statement analysis
St 1: a^x+y = 1. If a =1, x+y could be any number else it would be 0. INSUFFICIENT
St 2: b^x+y = 1. If b =1, x+y could be any number else it would be 0. INSUFFICIENT
St 1 & St 2: if a =1, b is not equal to 1. Hence x +y has to be 0 or even number if b = -1. Sufficient

Option C

EMPOWERgmatRichC · Answer

Hi All,

We're told that A and B are DISTINCT non-0 numbers. We're asked if X+Y is an EVEN integer. This is a YES/NO question that can be solved by TESTing VALUES.

1) (A^X)(A^Y) = 1

To start, it's important to note that we can 'rewrite' this equation as (A)^(X+Y) = 1

IF... 
A=1, then X and Y can be ANY VALUES. Thus, it's possible that X+Y could be an even integer (a 'YES' answer), an odd integer (a 'NO' answer) or a non-integer (also a 'NO' answer). 
Fact 1 is INSUFFICIENT

2) (B^X)(B^Y) = 1

The same logic that applies to Fact 1 also applies to Fact 2. We can rewrite this equation as (B)^(X+Y) = 1

IF... 
B=1, then X and Y can be ANY VALUES. Thus, it's possible that X+Y could be an even integer (a 'YES' answer), an odd integer (a 'NO' answer) or a non-integer (also a 'NO' answer). 
Fact 2 is INSUFFICIENT

Combined, we know that that A and B must be DISTINCT (meaning 'different') so while one of those variables could equal 1, the other would have to equal something else. Thus, we have to focus on that 'non 1' value and see what could happen...

IF... 
A = -1, then X and Y would have to either be BOTH EVEN or BOTH ODD. (Even + Even = Even, so the answer is 'YES') and (Odd + Odd = Even, so the answer is also 'YES'). 
A = anything integer other than -1, 0, or 1, then (X+Y) would have to equal 0 (since any non-0 value raised to the '0 power' equals 1) and the answer would also be 'YES'). 
The answer to the question is ALWAYS YES. 
Combined, SUFFICIENT

Final Answer:  C

GMAT assassins aren't born, they're made, 
Rich

Basshead · Answer

If a and b are distinct non zero numbers, is x+y an even integer?

(1)  a^x*a^y = 1 .

x + y can be odd. Insufficient.

(2)  b^x*b^y = 1

x + y can be odd. Insufficient.

(1+2) If a and b are distinct non-zero numbers, then x+y MUST be even.

a^x*a^y = 1^0*1^0 = 1

b^x*b^y = -1^0*-1^0 = 1

a^x*a^y = 1^2*1^2 = 1

b^x*b^y = -1^2*-1^2 = 1

Answer is C.

If a and b are distinct non zero numbers, is x+y an even

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