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M23-21

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Bunuel
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M23-21 [#permalink] New post   16 Sep 2014, 01:19
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81% (01:14) correct 19% (01:46) wrong based on 127 sessions
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How many integers \(x\) are there between 21 and 51, inclusive, such that \(x^3\) is divisible by both 3 and 5?

A. 1
B. 2
C. 3
D. 4
E. 5
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B
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Bunuel
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Re M23-21 [#permalink] New post   16 Sep 2014, 01:19
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Official Solution:

How many integers \(x\) are there between 21 and 51, inclusive, such that \(x^3\) is divisible by both 3 and 5?

A. 1
B. 2
C. 3
D. 4
E. 5


If \(x^3\) is divisible by 3 and 5, then \(x\) is also divisible by 3 and 5, and vice versa. We are looking for integers divisible by both 3 and 5 (i.e., divisible by 15). Between 21 and 51, there are 2 multiples of 15: 30 and 45.


Answer: B
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AB1607
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Re: M23-21 [#permalink] New post   06 Sep 2023, 12:48
What about 15? 15 also is a multiple of 15? so shouldn't the answer be 3- 15, 30 and 45?
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Re: M23-21 [#permalink] New post   06 Sep 2023, 13:00
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AB1607 wrote:
What about 15? 15 also is a multiple of 15? so shouldn't the answer be 3- 15, 30 and 45?


Hope the highlighted part will clear your doubt:

How many integers \(X\) are there between 21 and 51 inclusive such that \(X^3\) is divisible by both 3 and 5 ?
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Re: M23-21 [#permalink] New post   06 Sep 2023, 15:51
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M23-21 [#permalink]
06 Sep 2023, 15:51
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