GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
If k is a multiple of 24 but not a multiple of 16, which of
[#permalink]
12 Aug 2018, 15:36
12 Aug 2018, 15:36
1
Expert Reply
8
Bookmarks
00:00
Question Stats:
45% (01:26) correct
54% (01:58) wrong based on 96 sessions
Hide Show timer Statistics
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
[#permalink]
15 Aug 2018, 08:12
15 Aug 2018, 08:12
3
Expert Reply
2
Bookmarks
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 \(k = 81 \times 24.\)
General Discussion
Re: If k is a multiple of 24 but not a multiple of 16, which of
[#permalink]
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.
