M30-05

Official Solution:


If \(x\) and \(y\) are integers greater than 1, is \(x^y\) greater than 8?

(1) The sum of any two positive factors of \(x^2\) is even.

This statement implies that each factor of \(x^2\) is odd. That is because, if any factor were even, the sum of that factor and 1 would be odd. Next, if all factors of \(x^2\) are odd, then \(x^2\) is odd. For \(x^2\) to be odd, \(x\) must also be odd. The minimum value of \(x\) is then 3, and since \(y\) is greater than 1, the minimum value of \(y\) is 2. Consequently, the minimum value of \(x^y = 3^2 = 9\), which is greater than 8. Sufficient.

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

This statement implies that each factor of \(y^3\) is odd. That is because, if any factor were even, the product of that factor and another would be even. Next, if all factors of \(y^3\) are odd, then \(y^3\) is odd. For \(y^3\) to be odd, \(y\) must also be odd. The minimum value of \(y\) is then 3, and since \(x\) is greater than 1, the minimum value of \(x\) is 2. Consequently, the minimum value of \(x^y = 2^3 = 8\), which would give a NO answer to the question. However, if \(y\) is an odd number greater than 3, then \(x^y\) will be greater than 8, which would give a YES answer to the question. Not sufficient.


Answer: A
_________________

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.
_________________

(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

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.
_________________

I think this is a high-quality question and I agree with explanation.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
_________________