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If p is a positive integer and
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08 Sep 2021, 10:40
08 Sep 2021, 10:40
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If p is a positive integer and \(p^2\) is divisible by 12, then the largest positive integer that must divide \(p^3\) is
(A) \(2^3\)
(B) \(2^6\)
(C) \(3^3\)
(D) \(6^3\)
(E) \(12^2\)
(A) \(2^3\)
(B) \(2^6\)
(C) \(3^3\)
(D) \(6^3\)
(E) \(12^2\)
If p is a positive integer and
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08 Sep 2021, 22:48
08 Sep 2021, 22:48
1
Expert Reply
Given the statement, P must be an integer such that \(P^2\) is divisible by 12.
Let's start by writing the multiples of 12 to see if we get a perfect square.
12, 24, 36, 48,....
Notice that 36 is a perfect square of 6. Hence P=6.
Then, \(P^3=6^3\). Hence, the answer is D.
Let's start by writing the multiples of 12 to see if we get a perfect square.
12, 24, 36, 48,....
Notice that 36 is a perfect square of 6. Hence P=6.
Then, \(P^3=6^3\). Hence, the answer is D.
Re: If p is a positive integer and
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09 Sep 2021, 08:59
09 Sep 2021, 08:59
1
Since \(p^2\) is divisible by 12, which means by prime factorization, \(12 = 2^2 * 3\). so \(P^2\) is also divisible by \(2^2 * 3^2\), since there can't be odd powered when P is squared.
So for \(p^3\), it should be \(2^3 * 3^3 = 6^3\).
So for \(p^3\), it should be \(2^3 * 3^3 = 6^3\).
