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M30-05

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M30-05  [#permalink]

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New post 16 Sep 2014, 01:45
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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.

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Re M30-05  [#permalink]

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New post 16 Sep 2014, 01:45
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Official Solution:


(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
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Re: M30-05  [#permalink]

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New post 19 Oct 2018, 22:50
Hi Bunuel,

Could you please help me understand this part....."From this statement it follows that each factor of X square 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."

Since even + even = even AND odd + odd = even, why cannot we assume that all factors of X are even OR all factors of X are odd. In the first case lowest X would be 2(4)=8 and lowest Y is 2, therefore 8^2 > 8. Answer to question is Yes. In the 2nd case, lowest X would be 1(3)=3 and lowest Y = 2, therefore 3^2 >8. Answer to question is Yes. Sufficient.

Is my working wrong?
Thanks for your help.
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Re: M30-05  [#permalink]

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New post 20 Oct 2018, 02:02
achatterjee02 wrote:
Hi Bunuel,

Could you please help me understand this part....."From this statement it follows that each factor of X square 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."

Since even + even = even AND odd + odd = even, why cannot we assume that all factors of X are even OR all factors of X are odd. In the first case lowest X would be 2(4)=8 and lowest Y is 2, therefore 8^2 > 8. Answer to question is Yes. In the 2nd case, lowest X would be 1(3)=3 and lowest Y = 2, therefore 3^2 >8. Answer to question is Yes. Sufficient.

Is my working wrong?
Thanks for your help.


An integer cannot have all its factors even because 1 is a factor of every integer so at least one factor of very integer is odd.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M30-05  [#permalink]

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New post 03 Jan 2019, 20:53
(1) The sum of ANY two factors of \(x^2\) is even.

Let x be 3. So \(x^2 = 9\). Sum of any two factors of 9 is even. The minimum value x can take is 3. As y>1, \(x^y = 3^y\) and the minimum value of \(x^y\) is 9 which is greater than 8.
SUFFICIENT.

(2) The product of ANY two factors of \(y^3\) is odd.

Let y=3, \(y^3 = 27\). The product of any two factors of y is odd.
\(x^y = x^3\)
Now if x = 2, then \(x^y = 8\). The answer to the question is NO
If x = 3, \(x^y > 8\). The answer to the question is YES.
INSUFFICIENT.

OPTION: A
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Re: M30-05   [#permalink] 03 Jan 2019, 20:53
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