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30 Mar 2024, 22:38
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Question Stats:
100% (00:33) correct
0% (00:00) wrong based on 4 sessions
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Which of the following could be the units digit of \(57^n\), where \(n\) is a positive integer?
Indicate all such digits.
A. 0
B. 1
C. 2
D. 3
E. 4
F. 5
G. 6
H. 7
I. 8
J. 9
Indicate all such digits.
A. 0
B. 1
C. 2
D. 3
E. 4
F. 5
G. 6
H. 7
I. 8
J. 9
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Official Answer
B,D,H,J
12 Apr 2024, 16:19
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We have the number as \(57^n=(50+7)^n\), the units digit of the expansion will only depend on the units digit of the base 57.
for example 11^3 or 21^3 or 6781^3 will all have 1 as the units digit...\(11^1=11.... 11^2=11*11=121 ...11^3=121*11= 1331.\). thus \(11^1, 11^2, 11^3\) all have 1 as units digit,.
Thus 57^n will depend on 7...
Now the units digit have a cyclicity ..
0, 1, 5, 6 has a cyclicity of one that is with whatever power, they will have the same units digit..
4, 9 will have a cycilcity of 2, that is 4^1, 4^3, 4^5.. will have same units digit 4 and 4^2, 4^4, 4^6 will have 6 as units digit, that is 4, 6, 4, 6, .... 9 similarly will have 9, 1, 9, 1,...
2, 3, 7, 8 will have a cyclicity of 4...
2 -----\(2^1, 2^2, 2^3, 2^4, 2^5,2^6, 2^7,2^8\) will have units digit as 2, 4, 8, 6, 2, 4, 8, 6, ..
3-----3, 9, 7, 1, 3, 9, 7, 1, ....
7-----7, 9, 3, 1, 7, .....
8---- 8, 4, 2, 6, 8, 4, 2, 6, ....
so units digit when 7 is the units digit of base is 7, 9, 3, 1
for example 11^3 or 21^3 or 6781^3 will all have 1 as the units digit...\(11^1=11.... 11^2=11*11=121 ...11^3=121*11= 1331.\). thus \(11^1, 11^2, 11^3\) all have 1 as units digit,.
Thus 57^n will depend on 7...
Now the units digit have a cyclicity ..
0, 1, 5, 6 has a cyclicity of one that is with whatever power, they will have the same units digit..
4, 9 will have a cycilcity of 2, that is 4^1, 4^3, 4^5.. will have same units digit 4 and 4^2, 4^4, 4^6 will have 6 as units digit, that is 4, 6, 4, 6, .... 9 similarly will have 9, 1, 9, 1,...
2, 3, 7, 8 will have a cyclicity of 4...
2 -----\(2^1, 2^2, 2^3, 2^4, 2^5,2^6, 2^7,2^8\) will have units digit as 2, 4, 8, 6, 2, 4, 8, 6, ..
3-----3, 9, 7, 1, 3, 9, 7, 1, ....
7-----7, 9, 3, 1, 7, .....
8---- 8, 4, 2, 6, 8, 4, 2, 6, ....
so units digit when 7 is the units digit of base is 7, 9, 3, 1
23 Jun 2024, 11:25
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Official Explanation
The units digit of \(57^n\) is the same as the units digit of \(7^n\) for all positive integers n. To see why this is true for \(n = 2\), compute \(57^2\) by hand and observe how its units digit results from the units digit of \(7^2\). Because this is true for every positive integer n, you need to consider only powers of 7. Beginning with n = 1 and proceeding consecutively, the units digits of \(7, 7^2, 7^3, 7^4\), and \(7^5\) are 7, 9, 3, 1, and 7, respectively. In this sequence, the first digit, 7, appears again, and the pattern of four digits, 7, 9, 3, 1, repeats without end. Hence, these four digits are the only possible units digits of \(7^n\) and therefore of \(57^n\).
The correct answer consists of Choices B (1), D (3), H (7), and J (9).
The units digit of \(57^n\) is the same as the units digit of \(7^n\) for all positive integers n. To see why this is true for \(n = 2\), compute \(57^2\) by hand and observe how its units digit results from the units digit of \(7^2\). Because this is true for every positive integer n, you need to consider only powers of 7. Beginning with n = 1 and proceeding consecutively, the units digits of \(7, 7^2, 7^3, 7^4\), and \(7^5\) are 7, 9, 3, 1, and 7, respectively. In this sequence, the first digit, 7, appears again, and the pattern of four digits, 7, 9, 3, 1, repeats without end. Hence, these four digits are the only possible units digits of \(7^n\) and therefore of \(57^n\).
The correct answer consists of Choices B (1), D (3), H (7), and J (9).
