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16 Sep 2014, 00:43
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For a certain sequence \(a_1=9\) and \(a_n=(10^n - 1)^n\), where \(n\) is a positive integer. If S is the sum of the first \(n\) terms of the sequence, what is the units digit of S?
(1) \(n\) is even
(2) \(n\) is prime
(1) \(n\) is even
(2) \(n\) is prime
16 Sep 2014, 00:43
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Official Solution:
For a certain sequence \(a_1=9\) and \(a_n=(10^n - 1)^n\), where \(n\) is a positive integer. If S is the sum of the first \(n\) terms of the sequence, what is the units digit of S?
The terms of the sequence are given by the formula \(a_n=(10^n - 1)^n\).
\(a_1=9\)
\(a_2=99^2=...1\)
\(a_3=999^3=...9\)
\(a_4=9999^4=...1\)
...
As we can see odd terms have the units digit of 9 and the even terms have the units digit of 1.
The question asks about the units digit of the sum of the first \(n\) terms of the sequence. Now, if \(n\) is even then this sum ends with 0. If it's odd, then this sum ends with 9.
For example:
If \(n = 2\), then the units digit of the sum of 2 terms is \(9^1 + 99^2 = ...0\).
If \(n = 3\), then the units digit of the sum of 3 terms is \(9^1 + 99^2 + 999^3 = ...9\).
(1) \(n\) is even. As discussed above, if \(n\) is even, the sum ends with 0. Sufficient.
(2) \(n\) is prime.
All primes are odd except 2, so \(n\) can be even prime 2 or any odd prime. Hence, we cannot determine the units digit of S. Not sufficient.
Answer: A
For a certain sequence \(a_1=9\) and \(a_n=(10^n - 1)^n\), where \(n\) is a positive integer. If S is the sum of the first \(n\) terms of the sequence, what is the units digit of S?
The terms of the sequence are given by the formula \(a_n=(10^n - 1)^n\).
\(a_1=9\)
\(a_2=99^2=...1\)
\(a_3=999^3=...9\)
\(a_4=9999^4=...1\)
...
As we can see odd terms have the units digit of 9 and the even terms have the units digit of 1.
The question asks about the units digit of the sum of the first \(n\) terms of the sequence. Now, if \(n\) is even then this sum ends with 0. If it's odd, then this sum ends with 9.
For example:
If \(n = 2\), then the units digit of the sum of 2 terms is \(9^1 + 99^2 = ...0\).
If \(n = 3\), then the units digit of the sum of 3 terms is \(9^1 + 99^2 + 999^3 = ...9\).
(1) \(n\) is even. As discussed above, if \(n\) is even, the sum ends with 0. Sufficient.
(2) \(n\) is prime.
All primes are odd except 2, so \(n\) can be even prime 2 or any odd prime. Hence, we cannot determine the units digit of S. Not sufficient.
Answer: A
28 Feb 2023, 12:20
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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.
