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16 Apr 2008, 09:51
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If N is a natural integer, what is the remainder when 2 + 2^(8N+3) is divided by 5?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
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16 Apr 2008, 10:32
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Actually, we should find an unit digit of the expression.
Consider some powers of 2:
2^0: 1
2^1: 2
2^2: 4
2^3: 8
2^4: 16
2^5: 32
2^6: 64
2^7: 128
2^0: 256
2^0: 512
So, 2^x has a period of 4.
unit digit of 2^(8N+3) = unit digit of 2^(8*0+3) = unit digit of 2^3 = 8
unit digit of (2 + 8) = 0
Therefore, reminder is 0.
Consider some powers of 2:
2^0: 1
2^1: 2
2^2: 4
2^3: 8
2^4: 16
2^5: 32
2^6: 64
2^7: 128
2^0: 256
2^0: 512
So, 2^x has a period of 4.
unit digit of 2^(8N+3) = unit digit of 2^(8*0+3) = unit digit of 2^3 = 8
unit digit of (2 + 8) = 0
Therefore, reminder is 0.
16 Apr 2008, 13:38
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Factor out 2:
[2 (1 + 1^(8N+3))] / [5]
1 raised to any power is 1...therefore [2 (1+1)] / [5] = 4/5
Remainder when 4 is divided by 5 is 0. A
[2 (1 + 1^(8N+3))] / [5]
1 raised to any power is 1...therefore [2 (1+1)] / [5] = 4/5
Remainder when 4 is divided by 5 is 0. A
