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

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Manager
Joined: 06 Apr 2010
Posts: 139
If n is a positive integer, what is the remainder when [#permalink]

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23 Jun 2010, 11:35
1
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Question Stats:

78% (00:39) correct 22% (01:15) wrong based on 355 sessions

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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
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 43851
Re: PS Questions from SET 1-Need detail solution [#permalink]

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23 Jun 2010, 13:40
1
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udaymathapati wrote:
Some difficult PS Questions from SET1. Please help me with resolving them...OAs have mentioned at the end.

Q13:
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

3 in power has cyclicity of 4:
1. 3^1=3 (last digit is 3)
2. 3^2=9 (last digit is 9)
3. 3^3=27 (last digit is 7)
4. 3^4=81 (last digit is 1)

5. 3^5=243 (last digit is 3 again!)
...

To find the last digit of $$3^{8n+3}$$, divide the power (which is $$8n+3$$) by cyclicity # ( which is $$4$$) and look at the remainder --> $$\frac{8n+3}{4}$$ --> $$remainder=3$$, which means that the last digit of $$3^{8n+3}$$ will be the same as the last digit of $$3^3=27$$ (last digit is 7). (Side note: If the remainder were 0, then last digit would be the same as the las digit of $$3^4$$).

Now, last digit of $$3^{8n+3}+2$$ will be $$7+2=9$$. Any integer with last digit 9 upon division by 5 yields remainder of 4.

Hope it's clear.
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Joined: 06 Apr 2010
Posts: 139
Re: PS Questions from SET 1-Need detail solution [#permalink]

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23 Jun 2010, 20:35
Bunuel wrote:
udaymathapati wrote:
Some difficult PS Questions from SET1. Please help me with resolving them...OAs have mentioned at the end.

Q13:
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

3 in power has cyclicity of 4:
1. 3^1=3 (last digit is 3)
2. 3^2=9 (last digit is 9)
3. 3^3=27 (last digit is 7)
4. 3^4=81 (last digit is 1)

5. 3^5=243 (last digit is 3 again!)
...

To find the last digit of $$3^{8n+3}$$, divide the power (which is $$8n+3$$) by cyclicity # ( which is $$4$$) and look at the remainder --> $$\frac{8n+3}{4}$$ --> $$remainder=3$$, which means that the last digit of $$3^{8n+3}$$ will be the same as the last digit of $$3^3=27$$ (last digit is 7). (Side note: If the remainder were 0, then last digit would be the same as the las digit of $$3^4$$).

Now, last digit of $$3^{8n+3}+2$$ will be $$7+2=9$$. Any integer with last digit 9 upon division by 5 yields remainder of 4.

Hope it's clear.

Bunuel, : If the remainder were 0, then last digit would be the same as the las digit of 3^4=81 means 1.
Math Expert
Joined: 02 Sep 2009
Posts: 43851
Re: PS Questions from SET 1-Need detail solution [#permalink]

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23 Jun 2010, 21:01
1
KUDOS
Expert's post
udaymathapati wrote:

Bunuel, : If the remainder were 0, then last digit would be the same as the las digit of 3^4=81 means 1.

No, as in this case all numbers in power which is multiple of cyclicity would have the last digit of 1.

When power is divisible by cyclicity # (remainder 0), then the last digit is the same as the last digit of number in the power of cyclicity #.

For example:
Cyclicity of 2 is 4, so $$2^{4n}$$ (where $$n$$ is a positive integer) will have the same last digit as $$2^4$$, which is 6.

OR:

Cyclicity of 4 is 2, so $$4^{2n}$$ (where $$n$$ is a positive integer) will have the same last digit as $$4^2$$, which is 6.

Hope it's clear.
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Manager
Joined: 06 Apr 2010
Posts: 139
Re: PS Questions from SET 1-Need detail solution [#permalink]

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24 Jun 2010, 11:08
1
KUDOS
Bunuel wrote:
udaymathapati wrote:

Bunuel, : If the remainder were 0, then last digit would be the same as the las digit of 3^4=81 means 1.

No, as in this case all numbers in power which is multiple of cyclicity would have the last digit of 1.

When power is divisible by cyclicity # (remainder 0), then the last digit is the same as the last digit of number in the power of cyclicity #.

For example:
Cyclicity of 2 is 4, so $$2^{4n}$$ (where $$n$$ is a positive integer) will have the same last digit as $$2^4$$, which is 6.

OR:

Cyclicity of 4 is 2, so $$4^{2n}$$ (where $$n$$ is a positive integer) will have the same last digit as $$4^2$$, which is 6.

Hope it's clear.

Thanks bunuel. It's clear for me.
Intern
Joined: 22 Jun 2010
Posts: 2
Re: PS Questions from SET 1-Need detail solution [#permalink]

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25 Jun 2010, 20:03
Ok, beautiful solution, but to time consuming...
It´s a problem solving question, so, there is just one answer right?
Just to n = 1 and make the calculation...
Math Expert
Joined: 02 Sep 2009
Posts: 43851
Re: If n is a positive integer, what is the remainder when [#permalink]

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09 Mar 2014, 12:09
Bumping for review and further discussion.

For more on this kind of questions check Units digits, exponents, remainders problems collection.
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Posts: 13815
Re: If n is a positive integer, what is the remainder when [#permalink]

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09 Dec 2015, 19:02
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Location: India
GMAT 1: 650 Q45 V31
GPA: 4
Re: If n is a positive integer, what is the remainder when [#permalink]

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21 Aug 2017, 09:57
600 Level Question.
Took me 1 minute to solve.
Director
Joined: 07 Dec 2014
Posts: 906
If n is a positive integer, what is the remainder when [#permalink]

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21 Aug 2017, 11:36
udaymathapati 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

let n=1
check the units digits of 3,9,27,81
3^11, or 3^3+4+4, falls into the 3rd position in a cycle of 4, giving a units digit of 7
7+2=9
9/5 gives a remainder of 4
E
If n is a positive integer, what is the remainder when   [#permalink] 21 Aug 2017, 11:36
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