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

1. a^xa^y = 1. 2. b^xb^y = 1.

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.

Re: If a and b are distinct non zero numbers, is x+y an even [#permalink]

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26 Aug 2014, 10:55

Bunuel wrote:

surendar26 wrote:

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

1. a^xa^y = 1. 2. b^xb^y = 1.

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.

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.
_________________

Re: If a and b are distinct non zero numbers, is x+y an even [#permalink]

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05 May 2015, 01:18

Bunuel wrote:

surendar26 wrote:

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

1. a^xa^y = 1. 2. b^xb^y = 1.

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.

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

Last edited by ssriva2 on 05 May 2015, 02:03, edited 1 time in total.

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

1. a^xa^y = 1. 2. b^xb^y = 1.

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.

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=1 or a=b=1 and x+y=2=1

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.

Re: If a and b are distinct non zero numbers, is x+y an even [#permalink]

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23 Feb 2017, 05:09

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

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