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If n is a positive integer, what is the remainder when

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Manager
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If n is a positive integer, what is the remainder when [#permalink]

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New post 17 Jul 2008, 02:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Show the quickest way to solve this question.


Thks
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Re: Math Set 1: Q13 [#permalink]

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New post 17 Jul 2008, 03:41
3^(8n+3)+2 = 3^8n*3^3+2.
3 ^( anything to the form 4n where n=1,2,3,4) will end with 1.
(8n=4*2n)
3^3 ends with 7
so ....1*....7=no. ending with 7.
so (no. ending with 7) +2=no. ending with 9.
when divided by 5,remaider is 4

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Re: Math Set 1: Q13 [#permalink]

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New post 17 Jul 2008, 04:39
Another E.

arjtryarjtry has already given a great explanation!

arjtryarjtry wrote:
3^(8n+3)+2 = 3^8n*3^3+2.
3 ^( anything to the form 4n where n=1,2,3,4) will end with 1.
(8n=4*2n)
3^3 ends with 7
so ....1*....7=no. ending with 7.
so (no. ending with 7) +2=no. ending with 9.
when divided by 5,remaider is 4

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Re: Math Set 1: Q13 [#permalink]

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New post 17 Jul 2008, 05:50
In case you are not familiar with rules governing such powers, just write down last digits for powers of 3.

3^1 -- last digit =3
3^2 -- last digit = 9
3^3 -- last digit = 7
3^4 -- last digit = 1
3^5 -- last digit = 3 (starts repeating from here...)

so 8n+3 will always lead you to the 3 step in the iteration. 3,11,19..etc. so last digit 7+2 =9.
when divided by 5, remainder always is 4.

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Re: Math Set 1: Q13 [#permalink]

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New post 17 Jul 2008, 10:45
jimmylow wrote:
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

Show the quickest way to solve this question.


Thks


Powers of 3: 1 2 3 4
3 9 7 1 [unit digit]

for n=1,2,3,... we are looking for 3^11, 3^19 and 3^27. They all present unit digit of 7
7+2=9
9/5 => remainder = 4

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Re: Math Set 1: Q13 [#permalink]

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New post 06 Aug 2008, 16:48
arjtryarjtry wrote:
3^(8n+3)+2 = 3^8n*3^3+2.
3 ^( anything to the form 4n where n=1,2,3,4) will end with 1.
(8n=4*2n)
3^3 ends with 7
so ....1*....7=no. ending with 7.
so (no. ending with 7) +2=no. ending with 9.
when divided by 5,remaider is 4


This method works with a denominator of 5 with any number that ends with a 7 (as noted above):
(27+2)/5 = 5 remainder of 4
(37+2)/5 = 7 remainder of 4
(47+2)/5 = 9 remainder of 4 ---> all remainders of 4

but what if the denominator was 6:
(27+2)/6 = 4 remainder of 5
(37+2)/6 = 6 remainder of 3 ---> you get 2 different remainders. Am I missing something in the method above??
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Re: Math Set 1: Q13 [#permalink]

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New post 06 Aug 2008, 18:01
jimmylow wrote:
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

Show the quickest way to solve this question.


Thks

i took n=1, \(3^{8n+3} + 2\) = (27(3^8)+2)/5 = 5+((2(3^8)+2)/5)
3^8=6561 => 5+(((6561*2)+2 )/2)=> remainder = 4

Hence this is the method i used.Here there is no question of what can be the remainder its just what is the remnainder which means take any value of n and solve :)
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Re: Math Set 1: Q13 [#permalink]

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New post 07 Aug 2008, 11:03
jimmylow wrote:
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

Show the quickest way to solve this question.


Thks


the quickest, hm....

\(3^{1}\) = 3
\(3^{2}\) = 9
\(3^{3}\) = 27
\(3^{4}\) = 81
\(3^{5}\) = 243

see the pattern, so:

= \(3^{8n+3} + 2\)
= \(3^{8n} . 3^{3} + 2\)

since 8n is a multiple of 4 and therefore no matter the value of n, we can drop \(3^{8n}\)
8n:

= \(3^{3} + 2\)
= 29

so 4 is the reminder
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Re: Math Set 1: Q13   [#permalink] 07 Aug 2008, 11:03
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