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

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

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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

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Re: PS Questions from SET 1-Need detail solution [#permalink]

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New post 23 Jun 2010, 14:40
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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.

Answer: E.

Hope it's clear.
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Re: PS Questions from SET 1-Need detail solution [#permalink]

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New post 23 Jun 2010, 21:35
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.

Answer: E.

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.
shouldn't it be 3^0=1 instead?

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Re: PS Questions from SET 1-Need detail solution [#permalink]

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New post 23 Jun 2010, 22:01
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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.
shouldn't it be 3^0=1 instead?


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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Collection of Questions:
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: PS Questions from SET 1-Need detail solution [#permalink]

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New post 24 Jun 2010, 12:08
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.
shouldn't it be 3^0=1 instead?


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.

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Re: PS Questions from SET 1-Need detail solution [#permalink]

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New post 25 Jun 2010, 21:03
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...

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

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

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

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New post 21 Aug 2017, 10:57
600 Level Question.
Took me 1 minute to solve.

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

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New post 21 Aug 2017, 12:36
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

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If n is a positive integer, what is the remainder when   [#permalink] 21 Aug 2017, 12:36
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