RBBmba@2014 wrote:x,y are positive integers. Find the number of even factors of 4*x^2.
I. x^3-y^3+3xy is odd and x is a prime.
II. x^(x+y) * y^(3x) + x^(3y) is odd.
Since an ODD integer raised to a positive integer power stays ODD, and an EVEN integer raised to a positive integer power stays EVEN, the exponents in the two statements are irrelevant and can be disregarded.
Statement 1: x - y + 3xy is odd and x is a prime.
Case 1: y=1
Then
x - y + 3xy is odd becomes:
x - 1 + 3x = odd
4x = odd + 1
4x = even.
Case 1a: x=2
Here, 4x² = 16, which has even factors 2, 4, 8 and 16.
Case 1b: x=3
Here, 4x² = 36, which has even factors 2, 4, 6, 12, 18 and 36.
Since the number of even factors can be different values, INSUFFICIENT.
Statement 2: xy + x is odd
Case 1: y=1
Here,
xy + x = odd becomes:
2x = odd.
Not possible.
Case 2: y=2
Here,
xy + x = odd becomes:
3x = odd.
Here, it is possible that x=3 (Case 1b).
In Case 1b, 4x² = 36, which has even factors 2, 4, 6, 12, 18 and 36.
Case 2a: x=1
In this case, 4x² = 4, which has even factors 2 and 4.
Since the number of even factors can be different values, INSUFFICIENT.
Statements combined:
Only Case 1b satisfies both statements, implying that x must be an ODD PRIME NUMBER.
Thus, 4x² = 2²(odd prime)², with the result that 4x² will always yield the SAME NUMBER OF EVEN FACTORS.
SUFFICIENT.
The correct answer is
C.
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