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PS: remainder

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PS: remainder [#permalink]

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New post 24 Mar 2009, 23:44
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If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?
A. 0
B. 1
C. 2
D. 3
E. 4

Note: It is 3 raised to the power (8n+3).

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Re: PS: remainder [#permalink]

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New post 25 Mar 2009, 01:06
The answer is E..

Good explanations are provided below.

Last edited by shkusira on 26 Apr 2009, 08:01, edited 1 time in total.

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New post 25 Mar 2009, 03:07
I think Answer is E.

We look at the last digit for case divide by 5

3^(8n+3) + 2 = ((3^8)^n * 3^3) + 2 = ((6561)^n * 27) + 2 = ...7 + 2 = ...9

ps. (6561)^n <- the last digit is always 1.

from above the last digit is 9 so 9/5 has the remainder = 4.

Last edited by Takumi on 25 Mar 2009, 03:18, edited 1 time in total.

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New post 25 Mar 2009, 03:16
Agree with E.

if n=1
unit digit of 3 ^(8+3) is 7
unit digit of 3^11 + 2 is 9
when unit digit 9 divided by 5, the remainder is 4

consider whatever n is positive integer , unit digit of 3^(8n+3) is 7

then the answer is E.

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Re: PS: remainder [#permalink]

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New post 25 Mar 2009, 03:36
Last digit of 3^n will be 3,9,7,1 ( 3^1,3^2,3^3,3^4) and this cycle will repeat.

3^(8n+3) + 2 = ((3^8)^n * 3^3) + 2

(3^8) last digit will be 1 hence for (3^8)^n, last digit will be 1

Last digit for 3^3 is 7. Last digit for expression 3^(8n+3) will be 7

Hence last digit for 3^(8n+3) + 2 will be 9. If divided by 5 remainer will be 5.

irrespective of a digit at 10th place for this equation.

Cheers,

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Re: PS: remainder [#permalink]

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New post 25 Mar 2009, 14:27
Unit digit of 3^8n = 1
Unit digit of 3^3 = 7

So, unit digit of 3^8n x 3^3 = 7 --> unit digit +2 = 9 --> remainder is 4 when divided by 5 --> answer is E
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Re: PS: remainder [#permalink]

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New post 02 Jul 2009, 04:36
Ah!The great cycle theory!Wonderful.
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http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

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Re: PS: remainder   [#permalink] 02 Jul 2009, 04:36
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