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If N is a natural integer, what is the remainder when 2 +

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If N is a natural integer, what is the remainder when 2 + [#permalink]

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New post 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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Re: PS Integers [#permalink]

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New post 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.
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Re: PS Integers [#permalink]

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New post 16 Apr 2008, 13:38
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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Re: PS Integers [#permalink]

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New post 16 Apr 2008, 19:13
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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Re: PS Integers [#permalink]

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New post 16 Apr 2008, 23:36
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

Kudos [?]: 654 [0], given: 210

Senior Manager
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Re: PS Integers [#permalink]

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New post 17 Apr 2008, 10:03
bsd*,
Seems like 0 is not natural integer.
Natural integers are 1,2,3.......

Check this one out...
http://physicsmathforums.com/showthread.php?t=100

Have exam in 2 weeks...wonder what else I dont know :evil:

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Re: PS Integers [#permalink]

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New post 20 Apr 2008, 10:04
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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Re: PS Integers [#permalink]

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New post 20 Apr 2008, 19:09
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=100

Have exam in 2 weeks...wonder what else I dont know :evil:

Kudos [?]: 654 [0], given: 210

Re: PS Integers   [#permalink] 20 Apr 2008, 19:09
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