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27 Jun 2024, 01:30
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E
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Difficulty:
35%
(medium)
Question Stats:
73% (01:50) correct
27%
(02:06)
wrong
based on 120
sessions
History
Date
Time
Result
Not Attempted Yet
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 101 terms of the sequence, what is the units digit of S?
A. 0
B. 1
C. 3
D. 7
E. 9
A. 0
B. 1
C. 3
D. 7
E. 9
06 Jul 2024, 11:45
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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 101 terms of the sequence, what is the units digit of S?
A. 0
B. 1
C. 3
D. 7
E. 9
The terms of the sequence are given by the formula \(a_n = (10^n - 1)^n\).
\(a_1 = 9\)
\(a_2 = 99^2 = \dots 1\)
\(a_3 = 999^3 = \dots 9\)
\(a_4 = 9999^4 = \dots 1\)
...
As we can see, odd terms have the units digit of 9 and the even terms have the units digit of 1.
Now, if we add the first \(n\) terms where \(n\) is even, the sum ends with 0. If we add the first \(n\) terms where \(n\) is odd, this sum ends with 9.
For example:
If \(n = 2\), then the units digit of the sum of the first two terms is \(9 + 99^2 = 9+ ...1=\dots 0\).
If \(n = 3\), then the units digit of the sum of the first three terms is \(9 + 99^2 + 999^3 =9+...1+...9= \dots 9\).
The question asks about the units digit of the sum of the first 101, which is odd, terms of the sequence. Therefore, the units digit of \(S\) will be 9.
Answer: E
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 101 terms of the sequence, what is the units digit of S?
A. 0
B. 1
C. 3
D. 7
E. 9
The terms of the sequence are given by the formula \(a_n = (10^n - 1)^n\).
\(a_1 = 9\)
\(a_2 = 99^2 = \dots 1\)
\(a_3 = 999^3 = \dots 9\)
\(a_4 = 9999^4 = \dots 1\)
...
Now, if we add the first \(n\) terms where \(n\) is even, the sum ends with 0. If we add the first \(n\) terms where \(n\) is odd, this sum ends with 9.
For example:
If \(n = 2\), then the units digit of the sum of the first two terms is \(9 + 99^2 = 9+ ...1=\dots 0\).
If \(n = 3\), then the units digit of the sum of the first three terms is \(9 + 99^2 + 999^3 =9+...1+...9= \dots 9\).
Answer: E
General Discussion
27 Jun 2024, 01:52
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Solving for the first terms of the series
9,99^2,999^3,9999^4...
9 has a cyclicity of 2
so the digits are
9, ..1,..9,...1 etc which means every pair gives a 0 so 101st is 50 pairs -> 0 + 1st terms is 9
9,99^2,999^3,9999^4...
9 has a cyclicity of 2
so the digits are
9, ..1,..9,...1 etc which means every pair gives a 0 so 101st is 50 pairs -> 0 + 1st terms is 9
