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14 Feb 2014, 02:14
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\(x\) and \(y\) are two integers greater than 1. Is \(x^y\) greater than 8?
(1) The sum of ANY two factors of \(x^2\) is even.
(2) The product of ANY two factors of \(y^3\) is odd.
(1) The sum of ANY two factors of \(x^2\) is even.
(2) The product of ANY two factors of \(y^3\) is odd.
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14 Feb 2014, 02:15
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\(x\) and \(y\) are two integers greater than 1. Is \(x^y\) greater than 8?
(1) The sum of ANY two factors of \(x^2\) is even.
From this statement it follows that each factor of \(x^2\) is odd: if even one factor were even, the sum of at least one pair of factors, 1 and that even factor, would be odd. Next, if all the factors of \(x^2\) are odd, then \(x^2\) is odd. For \(x^2\) to be odd, \(x\) must also be odd. The least value of \(x\) is therefore 3 and since \(y\) is greater than 1, then the least value of \(y\) is 2. Thus the least value of \(x^y=3^2=9>8\). Sufficient.
(2) The product of ANY two factors of \(y^3\) is odd.
From this statement it follows that each factor of \(y^3\) is odd: if even one factor were even, the product of at least one pair of factors would be even. Next, if all the factors of \(y^3\) are odd, then \(y^3\) is odd. For \(y^3\) to be odd, \(y\) must also be odd. The least value of \(y\) is therefore 3 and since \(x\) is greater than 1, then the least value of \(x\) is 2. Thus the least value of \(x^y=2^3=8\), but if \(y=odd>3\), then \(x^y>8\). Not sufficient.
Answer: A.
(1) The sum of ANY two factors of \(x^2\) is even.
From this statement it follows that each factor of \(x^2\) is odd: if even one factor were even, the sum of at least one pair of factors, 1 and that even factor, would be odd. Next, if all the factors of \(x^2\) are odd, then \(x^2\) is odd. For \(x^2\) to be odd, \(x\) must also be odd. The least value of \(x\) is therefore 3 and since \(y\) is greater than 1, then the least value of \(y\) is 2. Thus the least value of \(x^y=3^2=9>8\). Sufficient.
(2) The product of ANY two factors of \(y^3\) is odd.
From this statement it follows that each factor of \(y^3\) is odd: if even one factor were even, the product of at least one pair of factors would be even. Next, if all the factors of \(y^3\) are odd, then \(y^3\) is odd. For \(y^3\) to be odd, \(y\) must also be odd. The least value of \(y\) is therefore 3 and since \(x\) is greater than 1, then the least value of \(x\) is 2. Thus the least value of \(x^y=2^3=8\), but if \(y=odd>3\), then \(x^y>8\). Not sufficient.
Answer: A.
General Discussion
14 Feb 2014, 04:57
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St1: The sum of ANY two factors of x^2 is even.
This means x is an odd integer. You can try by having x = 3, 5, 7, 9 etc. From stem, y can be 2 at a minimum. If x = 3 and y = 2, x^y = 3^2 = 9 so answer is yes to the stem question. Further, any combination of x and y will always have x^y > 8. Sufficient.
St2: The product of ANY two factors of y^3 is odd.
This means y is an odd integer. You can try by having y = 3, 5, 7, 9 etc. From stem, x can be 2 at a minimum. If x = 2 and y = 3, x^y = 2^3 = 8 so answer is no to the stem question. If x = 3 and y = 3, x^y = 3^3 = 27 so answer is yes to the stem question. Not sufficient.
Answer (C).
Answer (A). I hope I'm right.
p.s. I have edited my original answer.
This means x is an odd integer. You can try by having x = 3, 5, 7, 9 etc. From stem, y can be 2 at a minimum. If x = 3 and y = 2, x^y = 3^2 = 9 so answer is yes to the stem question. Further, any combination of x and y will always have x^y > 8. Sufficient.
St2: The product of ANY two factors of y^3 is odd.
This means y is an odd integer. You can try by having y = 3, 5, 7, 9 etc. From stem, x can be 2 at a minimum. If x = 2 and y = 3, x^y = 2^3 = 8 so answer is no to the stem question. If x = 3 and y = 3, x^y = 3^3 = 27 so answer is yes to the stem question. Not sufficient.
Answer (A). I hope I'm right.
p.s. I have edited my original answer.
