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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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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:

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