Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Quant Quizzes are back with a Bang and with lots of Prizes. The first Quiz will be on 8th Dec, 6PM PST (7:30AM IST). The Quiz will be Live for 12 hrs. Solution can be posted anytime between 6PM-6AM PST. Please click the link for all of the details.
Join IIMU Director to gain an understanding of DEM program, its curriculum & about the career prospects through a Q&A chat session. Dec 11th at 8 PM IST and 6:30 PST
(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.
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.
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.
(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.
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.
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.
close
41% (01:52) correct 
