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If n is a positive integer and n 2 is divisible by 96, then the larges
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15 Sep 2021, 01:19
15 Sep 2021, 01:19
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Question Stats:
80% (01:22) correct
20% (01:52) wrong based on 15 sessions
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If n is a positive integer and n^2 is divisible by 96, then the largest positive integer that must divide n is
(A) 6
(B) 12
(C) 24
(D) 36
(E) 48
(A) 6
(B) 12
(C) 24
(D) 36
(E) 48
Re: If n is a positive integer and n 2 is divisible by 96, then the larges
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15 Sep 2021, 09:09
15 Sep 2021, 09:09
1
If n^2 is divisible by 96, then first prime factorize 96, \(2^5 * 3\) which means n has at least \(2^3* 3 = 24\) as one of its prime factor.
About 36, prime factors are \(2^2 * 3^2\)there are 2 powers of 3 so may or may not divide n, so is case for 48.
About 36, prime factors are \(2^2 * 3^2\)there are 2 powers of 3 so may or may not divide n, so is case for 48.
Re: If n is a positive integer and n 2 is divisible by 96, then the larges
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25 Oct 2023, 14:46
25 Oct 2023, 14:46
eskay1981 wrote:
If n^2 is divisible by 96, then first prime factorize 96, \(2^5 * 3\) which means n has at least \(2^3* 3 = 24\) as one of its prime factor.
About 36, prime factors are \(2^2 * 3^2\)there are 2 powers of 3 so may or may not divide n, so is case for 48.
About 36, prime factors are \(2^2 * 3^2\)there are 2 powers of 3 so may or may not divide n, so is case for 48.
Sir, please clarify what does prime factor mean here? I thought prime factor meant prime numbers that are factors ...
