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16 Sep 2014, 00:17
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If \(x\) and \(y\) are positive integers and \(x\) is a prime factor of \(y^2\), then which of the following MUST be true?
A. \(x\) is even
B. \(x = y^2\)
C. \(\frac{y}{x}\) is even
D. \(\frac{x}{y} \le 1\)
E. \(x+1\) is also a prime factor of \(y^2\)
A. \(x\) is even
B. \(x = y^2\)
C. \(\frac{y}{x}\) is even
D. \(\frac{x}{y} \le 1\)
E. \(x+1\) is also a prime factor of \(y^2\)
16 Sep 2014, 00:17
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Official Solution:
If \(x\) and \(y\) are positive integers and \(x\) is a prime factor of \(y^2\), then which of the following MUST be true?
A. \(x\) is even
B. \(x = y^2\)
C. \(\frac{y}{x}\) is even
D. \(\frac{x}{y} \le 1\)
E. \(x+1\) is also a prime factor of \(y^2\)
Given that \(x\) is a prime factor of \(y^2\), and \(y\) is a positive integer, then \(x\) must be a factor of \(y\) as well - how else could \(x\) appear in the prime factorization of \(y^2\) (which is \(y*y\)) if it's not a factor of \(y\)? Accordingly, as \(x\) is a factor of \(y\) (and \(y\) is a positive integer), it follows that \(x \le y\). Therefore, \(\frac{x}{y} \le 1\), so option D must be true.
Alternative approach:
It's important to note that the question asks for what "MUST be true", not "COULD be true". In such questions, if we can demonstrate that a statement is NOT true for a specific set of numbers, it means that this statement is not always true, and therefore, not the correct answer.
A. \(x\) is even. This is not necessarily true. For example, if \(x=3\) (which is odd) and \(y=3\), then this statement doesn't hold.
B. \(x=y^2\). This is never true. As \(x\) is prime, it cannot be equal to the square of any integer.
C. \(\frac{y}{x}\) is even. This is not necessarily true. For instance, if \(x=3\) and \(y=3\), then \(\frac{y}{x}=1\), which is odd.
D. \(\frac{x}{y} \le 1\). This is always true, as explained above or via process of elimination.
E. \(x+1\) is also a prime factor of \(y^2\). This is not necessarily true. For example, if \(x=2\) and \(y=2\), then \(x+1=3\) and 3 is not a factor of \(y^2=4\).
Answer: D
If \(x\) and \(y\) are positive integers and \(x\) is a prime factor of \(y^2\), then which of the following MUST be true?
A. \(x\) is even
B. \(x = y^2\)
C. \(\frac{y}{x}\) is even
D. \(\frac{x}{y} \le 1\)
E. \(x+1\) is also a prime factor of \(y^2\)
Given that \(x\) is a prime factor of \(y^2\), and \(y\) is a positive integer, then \(x\) must be a factor of \(y\) as well - how else could \(x\) appear in the prime factorization of \(y^2\) (which is \(y*y\)) if it's not a factor of \(y\)? Accordingly, as \(x\) is a factor of \(y\) (and \(y\) is a positive integer), it follows that \(x \le y\). Therefore, \(\frac{x}{y} \le 1\), so option D must be true.
Alternative approach:
It's important to note that the question asks for what "MUST be true", not "COULD be true". In such questions, if we can demonstrate that a statement is NOT true for a specific set of numbers, it means that this statement is not always true, and therefore, not the correct answer.
A. \(x\) is even. This is not necessarily true. For example, if \(x=3\) (which is odd) and \(y=3\), then this statement doesn't hold.
B. \(x=y^2\). This is never true. As \(x\) is prime, it cannot be equal to the square of any integer.
C. \(\frac{y}{x}\) is even. This is not necessarily true. For instance, if \(x=3\) and \(y=3\), then \(\frac{y}{x}=1\), which is odd.
D. \(\frac{x}{y} \le 1\). This is always true, as explained above or via process of elimination.
E. \(x+1\) is also a prime factor of \(y^2\). This is not necessarily true. For example, if \(x=2\) and \(y=2\), then \(x+1=3\) and 3 is not a factor of \(y^2=4\).
Answer: D
02 May 2019, 15:40
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I picked a number to test out and found that to be easiest.
y=9, (y^2)=81, then x=3
A) x=even ; not true
B) x=(y^2); not true
C) (y/x)=even -> (9/3)=3; not true
D) (x/y)≤1 -> (3/9)= 1/3≤1; true
E) x+1 is also a prime factor of (y^2) -> 3+1=4; not true
Answer is D.
You can pick another set of numbers to test really quickly to see if you get the same answer, then move on.
If this helped you in any way, please send kudos!
y=9, (y^2)=81, then x=3
A) x=even ; not true
B) x=(y^2); not true
C) (y/x)=even -> (9/3)=3; not true
D) (x/y)≤1 -> (3/9)= 1/3≤1; true
E) x+1 is also a prime factor of (y^2) -> 3+1=4; not true
Answer is D.
You can pick another set of numbers to test really quickly to see if you get the same answer, then move on.
If this helped you in any way, please send kudos!
