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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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16 Sep 2014, 01:45
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: M3005
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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: M3005
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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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Re: M3005
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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: M3005
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12 Aug 2019, 19:47
Hi Bunuel,
if x=2, then x^2 = 4, factors are 1,2 4..then sum of any 2 factors is even . .2+4  6 even? and since y>1, y=2, so 2^2 = 4<8 makes stmt 1 insufficient? No. Please reply.



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Re: M3005
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12 Aug 2019, 22:32
san01sin wrote: Hi Bunuel,
if x=2, then x^2 = 4, factors are 1,2 4..then sum of any 2 factors is even . .2+4  6 even? and since y>1, y=2, so 2^2 = 4<8 makes stmt 1 insufficient? No. Please reply. (1) says that the sum of ANY two factors of x^2 is even. How is that true for 4? 1 + 2 = 3 = odd.










