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If x and y are positive integers, what is the remainder when y^x is

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If x and y are positive integers, what is the remainder when y^x is [#permalink]

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New post 24 Apr 2016, 10:54
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Re: If x and y are positive integers, what is the remainder when y^x is [#permalink]

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New post 24 Apr 2016, 19:30
St1: y^2 is odd --> y is odd
y^x = odd
odd/2 will always leave a remainder 1.
Sufficient

St2: xy is even --> y can be even or odd and remainder of (y^x)/2 can be 0 or 1.
Not Sufficient

Answer: A

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Re: If x and y are positive integers, what is the remainder when y^x is [#permalink]

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New post 30 Apr 2017, 08:42
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Bunuel wrote:
If x and y are positive integers, what is the remainder when y^x is divided by 2?

(1) y² is an odd integer.
(2) xy is an even integer.


Target question: What is the remainder when y^x is divided by 2?
NOTE: If we divide ANY positive integer by 2, the remainder will be EITHER 0 (if the number is even) OR 1 (if the number is odd)

Given: x and y are positive integers

Statement 1: y² is an odd integer
If y² is odd, then we know that y is ODD
If y is ODD, then y^x will be odd for ANY value of x (since x is a positive integer, y^x = the product of a bunch of y's)
If y^x is odd, then y^x divided by 2 will leave a remainder of 1
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: xy is an even integer
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 1 and y = 2. In this case, x^y = 1^2 = 1. So, when we divide x^y by 2, the remainder will be 1
Case b: x = 2 and y = 1. In this case, x^y = 2^1 = 2. So, when we divide x^y by 2, the remainder will be 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer:
[Reveal] Spoiler:
A


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Brent
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Re: If x and y are positive integers, what is the remainder when y^x is [#permalink]

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New post 24 Nov 2017, 14:14
Hi All,

We're told that X and Y are positive integers. We're asked for the remainder when Y^X is divided by 2.

This question can be solved by TESTing VALUES and/or by using Number Properties. It's worth noting that when dividing an integer by 2, the only possible remainders are 0 and 1.

1) Y^2 is an ODD integer.

Fact 1 tells us that Y^2 is an ODD integer - and we already know that X and Y are both POSITIVE INTEGERS.

(Even)^2 = Even
(Odd)^2 = Odd

This means that Y MUST be ODD. By extension, an ODD number raised to an INTEGER power will ALWAYS be ODD. Fact 1 essentially tells us that Y^X will ALWAYS be an ODD number. Dividing ANY odd number by 2 will ALWAYS give us a remainder of 1.
Fact 1 is SUFFICIENT

2) XY is an EVEN integer.

The information in Fact 2 means that one - or both - of the two integers are EVEN.

IF....
X=2, Y=1, then Y^X = 1 and the answer to the question is 1.
X=1, Y=2, then Y^X = 2 and the answer to the question is 0.
Fact 2 is INSUFFICIENT

Final Answer:
[Reveal] Spoiler:
A


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Kudos [?]: 3690 [0], given: 173

Re: If x and y are positive integers, what is the remainder when y^x is   [#permalink] 24 Nov 2017, 14:14
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If x and y are positive integers, what is the remainder when y^x is

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