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16 Sep 2014, 00:26
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If \(p\) is a prime number, is \(32p^3 - 16p^2 + 8p - 4\) divisible by \(p^2\)?
(1) \(p^2 < 25\)
(2) \(p^2-8p+12=0\)
(1) \(p^2 < 25\)
(2) \(p^2-8p+12=0\)
16 Sep 2014, 00:26
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Official Solution:
If \(p\) is a prime number, is \(32p^3 - 16p^2 + 8p - 4\) divisible by \(p^2\)?
Since the first two terms of \(32p^3 - 16p^2 + 8p - 4\) are divisible by \(p^2\), we need to determine if \(8p-4\) is divisible by \(p^2\).
(1) \(p^2 < 25\).
Since \(p\) is a prime number, it can be either \(p=2\) or \(p=3\). If \(p=2\), then \(8p-4\) is divisible by \(p^2=4\). However, if \(p=3\), then \(8p-4\) is NOT divisible by \(p^2=9\). Not sufficient.
(2) \(p^2-8p+12=0\).
Solving this equation, we find that \(p=2\) (the solution \(p=6\) is not valid since 6 is not a prime number). In the case of \(p=2\), we know that \(8p-4\) is divisible by \(p^2=4\). Sufficient.
Answer: B
If \(p\) is a prime number, is \(32p^3 - 16p^2 + 8p - 4\) divisible by \(p^2\)?
Since the first two terms of \(32p^3 - 16p^2 + 8p - 4\) are divisible by \(p^2\), we need to determine if \(8p-4\) is divisible by \(p^2\).
(1) \(p^2 < 25\).
Since \(p\) is a prime number, it can be either \(p=2\) or \(p=3\). If \(p=2\), then \(8p-4\) is divisible by \(p^2=4\). However, if \(p=3\), then \(8p-4\) is NOT divisible by \(p^2=9\). Not sufficient.
(2) \(p^2-8p+12=0\).
Solving this equation, we find that \(p=2\) (the solution \(p=6\) is not valid since 6 is not a prime number). In the case of \(p=2\), we know that \(8p-4\) is divisible by \(p^2=4\). Sufficient.
Answer: B
General Discussion
09 May 2023, 09:24
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
