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Which is an Integer?

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by dtweah » Thu May 28, 2009 7:27 am
If (X,Y) is a solution to 6x+8y=33 then which of the following three could be an interger?

I X^1/2

II PX, where P is prime

III QY, where Q is even

A. I only
B. II Only
C I and III only
D II and III only
E. I, II and III
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Re: Which is an Integer?

by Brent@GMATPrepNow » Thu May 28, 2009 7:58 am
dtweah wrote:If (X,Y) is a solution to 6x+8y=33 then which of the following three could be an interger?

I X^1/2

II PX, where P is prime

III QY, where Q is even

A. I only
B. II Only
C I and III only
D II and III only
E. I, II and III
If we want to know which of them "could" be an integer, then the answer is E.
The solution (4, 9/8) works for all three
1. 4^(1/2) = 2 (integer)
2. 4 multiplied by any prime number will yield an integer
3. if Q=8, then QY = 8(9/8)=9 (integer)
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by mikeCoolBoy » Thu May 28, 2009 8:02 am
IMO D

we can rewrite the expression as

6x + 8y = 33 --> 8y = 33 - 6x ----> 8y = 3(11-2x)

we can see that y = 3 and x = 1.5 are in the line so

PX where P is prime, choosing P = 2 makes PX = 3 which is integer.
QY where Q is even, choosing Q = 2 makes QY = 6 which is an integer.
So II and III could be true.

8y is an even number
so we need 11 - 2x to be an even number so that the product of 3 and 11-2x is an even number.

11-2x has to be even ---> so 2x has to be odd which is impossible unless x is not an integer and therefore x ^ (1/2) cannot be an integer.

Choose D.
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by mikeCoolBoy » Thu May 28, 2009 8:50 am
You're completely right Bren, I wrongly assumed that Y was an integer complicating the problem.
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by sana.noor » Tue Sep 17, 2013 11:20 pm
i can see that a single word "could" has changed the entire question. we can prove all three to be true. what if we restrict the value of x and y to be only integers, does this effect the final answer?
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