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12 Dec 2019, 03:09
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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:
15%
(low)
Question Stats:
76% (00:56) correct
24%
(01:07)
wrong
based on 215
sessions
History
Date
Time
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Not Attempted Yet
What is the remainder when \(3^{283}\) is divided by 5?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
12 Dec 2019, 03:24
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Bunuel
Explanation:
Go With cyclicity of 3, We know it is 4
3^1 = 3
3^2 = 9
3^3 = 27 (need last digit only)
3^4 = 81 (need last digit only)
power / 4= 283/4 = 3 order of remainder.
so, 7/5= 2 Remainder
IMO-C
15 Dec 2019, 09:56
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Alternatively,
(3^283)/5 = (3*(3^2)^141)/5= (3*(10-1)^141)/5 = 3*(-1/5) = -3/5 hence remainder = -3+5 = 2
The answer is C.
Posted from my mobile device
(3^283)/5 = (3*(3^2)^141)/5= (3*(10-1)^141)/5 = 3*(-1/5) = -3/5 hence remainder = -3+5 = 2
The answer is C.
Posted from my mobile device
