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Manager
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If N is a natural integer, what is the remainder when 2 + [#permalink]
 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
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CEO
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Concentration: Entrepreneurship, Other
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Actually, we should find an unit digit of the expression. Consider some powers of 2: 2^0: 1 2^1: 22^2: 4 2^3: 8 2^4: 16 2^5: 322^6: 64 2^7: 128 2^0: 256 2^0: 512So, 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.
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Intern
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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
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VP
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JCLEONES wrote: 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 Pick N=1 So you have 2 + 2^11 Knowing that 2^5 has digit of 2, the digit of 2^11 = 8 2+8 = 10 10/5 has remainder=0
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CEO
Joined: 17 May 2007
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You're kidding right ? Are you sure you can factor out 2 like that ? Also when 4 is divided by 5 , remainder is 4 not zero. tikpaklong wrote: 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
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Senior Manager
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bsd*, Seems like 0 is not natural integer. Natural integers are 1,2,3....... Check this one out... http://physicsmathforums.com/showthread.php?t=100Have exam in 2 weeks...wonder what else I dont know 
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Director
Joined: 14 Oct 2007
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since all the Answer choices are constants , we know that the remainder will be constant for ANY value of N.
so just put N = 1
2+2^11 = 2050 which means that units digit is 0 and therefore remainder has to be 0
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CEO
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Dont worry kyatin - if they ask about natural integer's you'll know that you're looking at 48+ in quant  kyatin wrote: bsd*, Seems like 0 is not natural integer. Natural integers are 1,2,3....... Check this one out... http://physicsmathforums.com/showthread.php?t=100Have exam in 2 weeks...wonder what else I dont know |
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