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12 Nov 2021, 08:45
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B
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Difficulty:
75%
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Question Stats:
50% (01:52) correct
50%
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wrong
based on 177
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Numbers A, B, C, and D have 16, 28, 30, and 27 factors, respectively. Which of these four numbers could be a perfect cube?
A. Only A
B. A and B
C. C and D
D. A, B, and C
E. A, B, C, and D
A. Only A
B. A and B
C. C and D
D. A, B, and C
E. A, B, C, and D
PyjamaScientist
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12 Nov 2021, 09:28
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Option (B).
Concept tested: If \(N = a^p× b^q × c^r\) …, then N has \((p+1)(q+1)(r+1)\)… number of factors.
16 = 4×4. Powers can be 3 and 3.
28 = 4×7. Powers can be 3 and 6.
You can't generate similar pairs for 30 and 27. Also, total no. of factors would be even for perfect cubes, so 27 can also be eliminated that way.
Concept tested: If \(N = a^p× b^q × c^r\) …, then N has \((p+1)(q+1)(r+1)\)… number of factors.
16 = 4×4. Powers can be 3 and 3.
28 = 4×7. Powers can be 3 and 6.
You can't generate similar pairs for 30 and 27. Also, total no. of factors would be even for perfect cubes, so 27 can also be eliminated that way.
andrenoble
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14 Nov 2021, 07:36
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PyjamaScientist
Hi!
Thank you for your post. I went the same path as you, my only concern is 28: if I understand correctly, since all of the numbers are perfect cubes, it's reasonable to re-write the eqaution as follows:
\(N = 1× a^3\), and thus the number of factors of a will be the deciding factor to solve the problem.
If a has 2 factors, we could rewrite the eqaution as \(N = 1× (a'^3) × (a"^2)\), where a' and a" denote factors of a. We will get 16 factors, hence A could be such number.
If a has 3 factors, we could rewrite the equation as \(N = 1× (a'^3) × (a"^2) × (a"'^2)\), and it gives us 64 factors of any perfect cube.
That's why I think it's only A, but please correct me if I'm wrong/missing something.
