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16 Sep 2014, 00:44
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If \(n\) is an non-negative integer, is \(10^n+8\) divisible by 18?
(1) \(n\) is a prime number.
(2) \(n\) is an even number.
(1) \(n\) is a prime number.
(2) \(n\) is an even number.
16 Sep 2014, 00:44
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Official Solution:
If \(n\) is an non-negative integer, is \(10^n+8\) divisible by 18?
Observe that \(10^n+8\) is divisible by 18 for any positive integer value of \(n\). In this case, \(10^n+8=\text{even}+\text{even}=\text{even}\), so it's divisible by 2. Additionally, the sum of the digits of \(10^n+8\) is 9, so it's divisible by 9. Since \(10^n+8\) is divisible by both 2 and 9, it is also divisible by \(2*9=18\).
However, if \(n=0\), then \(10^n+8=1+8=9\), so in this case, \(10^n+8\) is not divisible by 18.
Therefore, the question essentially asks whether \(n\) is a positive integer.
(1) \(n\) is a prime number.
Since \(n\) is a prime number, it must be a positive integer. Sufficient.
(2) \(n\) is an even number.
\(n\) can be zero as well as any positive even number. Not sufficient.
Answer: A
If \(n\) is an non-negative integer, is \(10^n+8\) divisible by 18?
Observe that \(10^n+8\) is divisible by 18 for any positive integer value of \(n\). In this case, \(10^n+8=\text{even}+\text{even}=\text{even}\), so it's divisible by 2. Additionally, the sum of the digits of \(10^n+8\) is 9, so it's divisible by 9. Since \(10^n+8\) is divisible by both 2 and 9, it is also divisible by \(2*9=18\).
However, if \(n=0\), then \(10^n+8=1+8=9\), so in this case, \(10^n+8\) is not divisible by 18.
Therefore, the question essentially asks whether \(n\) is a positive integer.
(1) \(n\) is a prime number.
Since \(n\) is a prime number, it must be a positive integer. Sufficient.
(2) \(n\) is an even number.
\(n\) can be zero as well as any positive even number. Not sufficient.
Answer: A
General Discussion
07 May 2023, 05:12
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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.
