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If n is a positive integer, what is the remainder when 38n+3

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Manager
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If n is a positive integer, what is the remainder when 38n+3 [#permalink] New post 31 Jan 2007, 05:17
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If n is a positive integer, what is the remainder when 38n+3 + 2 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4
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Kudos [?]: 83 [0], given: 39

 [#permalink] New post 31 Jan 2007, 08:56
tough guy , i beleive u need to repost the question and make it more clear ( the is 38^(n+3)+2 or is it 38n+3+2 ie : 38n+5 or is it

wut exactly

thanks
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 [#permalink] New post 31 Jan 2007, 17:45
not sure how to solve this..38^(n+3) +2 divided by 5, has remainders all over the place

i am only focusing on the unit digit of 38, which is determined by 8, unit digits are 2, 6, 4 etc +2 =4, 8, 6, which give remainder 4, 3, and 1
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 [#permalink] New post 01 Feb 2007, 00:07
If question is asking to find the remainder when 38^(n+3) +2 is divided by 5
then there is no single answer

Put n=1, then unit digits of 38^4 =6. then the unit's digit of the number =8.Therefore, the remainder would be 3

put n=2, unit digits of 38^5=8, then unit digits of the number=0. therefore there would be no remainder.
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 [#permalink] New post 01 Feb 2007, 09:11
Guys,

Sorry about the confusion.

The source is unknown and I beleive that the question is not very clear.

I will get back to you with more details

Tough guy
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  [#permalink] 01 Feb 2007, 09:11
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