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If k is a multiple of 24 but not a multiple of 16, which of
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12 Aug 2018, 15:36
12 Aug 2018, 15:36
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Question Stats:
49% (01:24) correct
50% (01:50) wrong based on 83 sessions
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If k is a multiple of 24 but not a multiple of 16, which of the following cannot be an integer?
(A) \(\frac{k}{8}\)
(B) \(\frac{k}{9}\)
(C) \(\frac{k}{32}\)
(D) \(\frac{k}{36}\)
(E) \(\frac{k}{81}\)
(A) \(\frac{k}{8}\)
(B) \(\frac{k}{9}\)
(C) \(\frac{k}{32}\)
(D) \(\frac{k}{36}\)
(E) \(\frac{k}{81}\)
Re: If k is a multiple of 24 but not a multiple of 16, which of
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15 Aug 2018, 08:12
15 Aug 2018, 08:12
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Explanation
If k is a multiple of 24, it contains the prime factors of 24: 2, 2, 2, and 3. (It could also contain other prime factors, but the only ones for certain are the prime factors contained in 24.)
If k were a multiple of 16, it would contain the prime factors of 16: 2, 2, 2, and 2.
Thus, if k is a multiple of 24 but not of 16, k must contain 2, 2, and 2, but not a fourth 2 (otherwise, it would be a multiple of 16).
Thus: k definitely has 2, 2, 2, and 3. It could have any other prime factors (including more 3’s) except for more 2’s.
An answer choice in which the denominator contains more than three 2’s would guarantee a non-integer result. Only choice (C) works. Since k has fewer 2’s than 32, can never be an integer.
Alternatively, list multiples of 24 for k: 24, 48, 72, 96, 120, 144, 168, etc. Then, eliminate multiples of 16 from this list: 24, , 72, , 120, , 168, etc.
A pattern emerges: k = (an odd integer) \(\times 24\):
(A) can be an integer, for example when \(k = 24.\)
(B) can be an integer, for example when \(k = 72.\)
(C) is correct by process of elimination.
(D) can be an integer, for example when \(k = 72.\)
(E) can be an integer, for example when \(k = 81 \times 24.\)
If k is a multiple of 24, it contains the prime factors of 24: 2, 2, 2, and 3. (It could also contain other prime factors, but the only ones for certain are the prime factors contained in 24.)
If k were a multiple of 16, it would contain the prime factors of 16: 2, 2, 2, and 2.
Thus, if k is a multiple of 24 but not of 16, k must contain 2, 2, and 2, but not a fourth 2 (otherwise, it would be a multiple of 16).
Thus: k definitely has 2, 2, 2, and 3. It could have any other prime factors (including more 3’s) except for more 2’s.
An answer choice in which the denominator contains more than three 2’s would guarantee a non-integer result. Only choice (C) works. Since k has fewer 2’s than 32, can never be an integer.
Alternatively, list multiples of 24 for k: 24, 48, 72, 96, 120, 144, 168, etc. Then, eliminate multiples of 16 from this list: 24, , 72, , 120, , 168, etc.
A pattern emerges: k = (an odd integer) \(\times 24\):
(A) can be an integer, for example when \(k = 24.\)
(B) can be an integer, for example when \(k = 72.\)
(C) is correct by process of elimination.
(D) can be an integer, for example when \(k = 72.\)
(E) can be an integer, for example when \(k = 81 \times 24.\)
Re: If k is a multiple of 24 but not a multiple of 16, which of
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18 Sep 2019, 03:34
18 Sep 2019, 03:34
2
Is there any alternative way to solve this but not like Manhattan 5lb or discuss in above.
