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Feb 11, 2020 6:18 am
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If n is a positive integer and n^4 is divisible by 720, which of the following may not always divide n^3?
A) 36
B) 80
C) 120
D) 200
E) 360
A) 36
B) 80
C) 120
D) 200
E) 360
Feb 17, 2020 1:41 pm
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720 = 8 * 9 * 10 = 2^4 * 3^2 * 5
So we know n^4 is divisible by 2^4 * 3^2 *5, which will be true as long as n is divisible by 2 * 3 * 5. So n^3 must be divisible by 2^3 * 3^3 * 5^3. We can't be sure that n^3 is divisible by 2^4, though, so we can't be sure n^3 is divisible by 80 = 2^4 * 5, and B is the right answer.
The wording of the question is strange though, since it first discusses n as if it represents a single unknown number, and at the end discusses n as if it's a variable. What is the source?
So we know n^4 is divisible by 2^4 * 3^2 *5, which will be true as long as n is divisible by 2 * 3 * 5. So n^3 must be divisible by 2^3 * 3^3 * 5^3. We can't be sure that n^3 is divisible by 2^4, though, so we can't be sure n^3 is divisible by 80 = 2^4 * 5, and B is the right answer.
The wording of the question is strange though, since it first discusses n as if it represents a single unknown number, and at the end discusses n as if it's a variable. What is the source?
Jul 08, 2024 9:10 am
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\(720 = 2^4*3^2*5^1\)
\(\frac{(n)^4}{2^4*3^2*5^1}\)
Therefore, \(n^4\) must at least contain four 2s, two 3s and one 5.
So \(n^4 = (2*3*5)^4\) satisifes our requirement. Because it has four 2s, four 3s and four 5s, above our minimum requirement.
Therefore, \(n^3 = (2*3*5)^3\). Take prime factorization of the options and you'll see that only option B is not divissible.
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If \(n^4\) would have been divided by \(2^5*3^2*5^1\), then \(n^4 = (2^2*3*5)^4\)
\(\frac{(n)^4}{2^4*3^2*5^1}\)
Therefore, \(n^4\) must at least contain four 2s, two 3s and one 5.
So \(n^4 = (2*3*5)^4\) satisifes our requirement. Because it has four 2s, four 3s and four 5s, above our minimum requirement.
Therefore, \(n^3 = (2*3*5)^3\). Take prime factorization of the options and you'll see that only option B is not divissible.
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If \(n^4\) would have been divided by \(2^5*3^2*5^1\), then \(n^4 = (2^2*3*5)^4\)
