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08 Dec 2022, 12:15
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
66% (02:02) correct
34%
(02:01)
wrong
based on 303
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Not Attempted Yet
If n and m are positive integers, what is the last digit of \(83^{56m} + (10^2 – 1)^{64n}\)?
A. 0
B. 1
C. 2
D. 3
E. 9
A. 0
B. 1
C. 2
D. 3
E. 9
08 Dec 2022, 13:33
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Bunuel
\(83^{56m} + (10^2 – 1)^{64n}\)
\(83^{56m} + (99)^{64n}\)
The cyclicity of 3 is 3, 9 , 7, 1
The cyclicity of 9 is 9, 1, 9, 1
56m & 64n divisible by 4, hence the unit digit = 1 + 1 = 2
Option C
Archit3110
Major Poster
09 Dec 2022, 02:28
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56 is multiple of 4 ; 56 * m will always be even
83^ multiple of 4 ; unit digit 1
(10^2-1) ^ 64 m
or say ( 99)^64m
9^ even power has unit digit 1
1+1 ; 2
option C
83^ multiple of 4 ; unit digit 1
(10^2-1) ^ 64 m
or say ( 99)^64m
9^ even power has unit digit 1
1+1 ; 2
option C
Bunuel
