Which of the following equations has only one integer pair as solution?
A) y = 2x
B) y = x/2
C) y=x √5
D) y=x+1
E) y=1/x
OA: C
Which of the following equations has
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For four of the five answer choices, show that there is more than one integer pair.NandishSS wrote:Which of the following equations has only one integer pair as solution?
A) y = 2x
B) y = x/2
C) y=x √5
D) y=x+1
E) y=1/x
The correct answer will be the remaining answer choice.
A: y = 2x
Case 1; x=1, y=2
Case 2: x=2, y=4
Since more than one integer pair is possible, eliminate A.
B: y = x/2
Case 1: x=2, y=1
Case 2: x=4, y=2
Since more than one integer pair is possible, eliminate B.
D: y = x+1
Case 1: x=1, y=2
Case 2: x=2, y=3
Since more than one integer pair is possible, eliminate D.
E: y = 1/x
Case 1: x=1, y=1
Case 2: x=-1, y=-1
Since more than one integer pair is possible, eliminate E.
The correct answer is C.
C: y = x√5
Her, the only integer pair is x=0, y=0.
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Hi NandishSS,
This is a real 'concept' question - meaning that if you understand the concept(s) involved, then you don't have to do much math to get to the solution. As it stands, Mitch has done a nice job of TESTing VALUES to prove which 4 answers have MORE than 1 set of integers solutions for X and Y (and that's a fairly efficient way of answering this question).
We're told that X and Y have to be INTEGERS. In most of the answer choices, we're either multiplying, dividing or adding elements - and that type of work in just one equation almost always produces multiple integer solutions. Ian brings up a good point below, so I'll be more specific with what I was trying to communicate: In Answer C, we're multiplying by a number with a non-repeating, non-terminating decimal - and that means that the only way to end with an integer is to multiply by 0 (since 0 multiplied by anything is 0). In simple terms, you cannot multiply the square-root of 5 by any other integer than 0 and get an integer result.
Final Answer: C
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This is a real 'concept' question - meaning that if you understand the concept(s) involved, then you don't have to do much math to get to the solution. As it stands, Mitch has done a nice job of TESTing VALUES to prove which 4 answers have MORE than 1 set of integers solutions for X and Y (and that's a fairly efficient way of answering this question).
We're told that X and Y have to be INTEGERS. In most of the answer choices, we're either multiplying, dividing or adding elements - and that type of work in just one equation almost always produces multiple integer solutions. Ian brings up a good point below, so I'll be more specific with what I was trying to communicate: In Answer C, we're multiplying by a number with a non-repeating, non-terminating decimal - and that means that the only way to end with an integer is to multiply by 0 (since 0 multiplied by anything is 0). In simple terms, you cannot multiply the square-root of 5 by any other integer than 0 and get an integer result.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
Last edited by [email protected] on Wed Jul 19, 2017 4:45 pm, edited 1 time in total.
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That is not correct. If you have this equation:[email protected] wrote: In Answer C though, we're multiplying by a number with a non-terminating decimal - and that means that the only way to end with an integer is to multiply by 0 (since 0 multiplied by anything is 0).
x = (1/3)*y
you're multiplying y by a non-terminating decimal, but you don't need something to end in zero to get integer solutions; you can have x = 1 and y = 3 for example.
The difficulty in C is that we're multiplying by √5, which is an irrational number (a number that cannot be written as a fraction with integer numerator and denominator). You can never multiply an irrational number by a nonzero integer and get an integer result, which is why 0 is the only value that can work in answer C. But that's not a fact test takers will ever need to know on the GMAT, and it would be unreasonable to expect test takers to prove that fact from scratch in two minutes. Since for every GMAT question there will always be a reasonable way to prove the right answer is correct (rather than just a way to rule out the four other wrong answers) this is not a realistic GMAT question.
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The question wants us to pick an option that renders only one integer value of x and only one compatible integer value of y.NandishSS wrote:Which of the following equations has only one integer pair as solution?
A) y = 2x
B) y = x/2
C) y=x √5
D) y=x+1
E) y=1/x
OA: C
Let's discuss all the options one by one.
A) y = 2x => It implies that y is a multiple of '2;' thus, there can be many compatible integer values of x and y. For example, (x, y): (0, 0); (1, 2); (-1, -2), etc.
B) y = x/2 => x = 2y => It implies that x is a multiple of '2;' thus, there can be many compatible integer values of x and y. For example, (x, y): (0, 0); (2, 1); (-2, -1), etc.
C) y = x √5 => One of the integer pairs for (x, y) is (0, 0). To get y an integer value, x must be a multiple of √5; however, this will not serve the purpose as we want integer pairs for (x, y). Thus, y = x √5 has only one integer pairs for (x, y) i.e. (0, 0).
D) y = x+1 => Many integer pairs of values for x and y are possible. For example, (x, y): (0, 1); (-1, 0); (1, 2), etc.
E) y = 1/x => xy = 1 => If the product of two integers is '1,' they each can be either 1 or -1. Thus, there are two integer pairs for the values of x and y, and they are (1, 1) ans (-1, -1).
The correct answer: C
Hope this helps!
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I'd go right to C: that √5 can only be turned into an integer by being multiplied by something else with a √5 in it (which wouldn't be an integer), or by 0. So x must be 0.
From there, y must also be 0, and the only integer solution to C is x = 0, y = 0.
From there, y must also be 0, and the only integer solution to C is x = 0, y = 0.
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Another way to get there:
A and B are two sides of the same coin (2x = y and 2y = x), so they'll both have the same number of solutions.
D has an infinite number of solutions: {1,2} and {2, 3} and {3,4} and ...
E is trickier, but xy = 1 has x = 1, y = 1 and x = -1, y = -1 as solutions, so we've got two pairs.
From there, C is the only possibility!
A and B are two sides of the same coin (2x = y and 2y = x), so they'll both have the same number of solutions.
D has an infinite number of solutions: {1,2} and {2, 3} and {3,4} and ...
E is trickier, but xy = 1 has x = 1, y = 1 and x = -1, y = -1 as solutions, so we've got two pairs.
From there, C is the only possibility!
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We see that choices A, B and D will have infinitely many integer pair solutions. Choice E has two, namely (1, 1) and (-1, -1). However, choice C only has one, namely, (0, 0).
Answer: C
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