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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.

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

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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?

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

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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...

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

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09 Dec 2015, 20:02

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

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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