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16 May 2024, 01:30
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D
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Difficulty:
55%
(hard)
Question Stats:
60% (01:23) correct
40%
(01:11)
wrong
based on 254
sessions
History
Date
Time
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Not Attempted Yet
If 11 is a factor of \(n^2\) and \(\sqrt{n}\) is a prime number, what is the sum of the digits of the positive integer \(n\)?
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
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02 Jun 2024, 05:00
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Official Solution:
If 11 is a factor of \(n^2\) and \(\sqrt{n}\) is a prime number, what is the sum of the digits of the positive integer \(n\)?
A. 1
B. 2
C. 3
D. 4
E. 5
Exponentiation does not "produce" primes. Since 11 is a factor of \(n^2\), it must also be a factor of \(n\). Essentially, this condition implies that \(n\) is a multiple of 11.
Furthermore, the condition that \(\sqrt{n}\) is a prime number implies \(n = \text{prime}^2\).
Given that \(n\) is a multiple of 11 and \(n = \text{prime}^2\), it follows that \(n = 11^2 = 121\). Therefore, the sum of the digits of \(n\) is \(1 + 2 + 1 = 4\).
Answer: D
If 11 is a factor of \(n^2\) and \(\sqrt{n}\) is a prime number, what is the sum of the digits of the positive integer \(n\)?
A. 1
B. 2
C. 3
D. 4
E. 5
Exponentiation does not "produce" primes. Since 11 is a factor of \(n^2\), it must also be a factor of \(n\). Essentially, this condition implies that \(n\) is a multiple of 11.
Furthermore, the condition that \(\sqrt{n}\) is a prime number implies \(n = \text{prime}^2\).
Given that \(n\) is a multiple of 11 and \(n = \text{prime}^2\), it follows that \(n = 11^2 = 121\). Therefore, the sum of the digits of \(n\) is \(1 + 2 + 1 = 4\).
Answer: D
General Discussion
16 May 2024, 02:24
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Bunuel
\(\sqrt{n}\) = 11
\(n = 11^2 = 121\)
\(n^2 = 11^4\)
The sum of the digits of the positive integer \(n\) = 1 + 2 + 1 = 4
Option D
