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If n is a positive integer, what is the remainder when
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23 Jun 2010, 11:35
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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
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Re: PS Questions from SET 1Need detail solution
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23 Jun 2010, 13:40
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 1Need detail solution
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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. 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 1Need detail solution
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23 Jun 2010, 21:01
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: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat 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 1Need detail solution
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24 Jun 2010, 11:08
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 1Need detail solution
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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...



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Re: If n is a positive integer, what is the remainder when
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09 Mar 2014, 12:09



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Re: If n is a positive integer, what is the remainder when
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21 Aug 2017, 09:57
600 Level Question. Took me 1 minute to solve.



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



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If n is a positive integer, what is the remainder when
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27 Sep 2018, 23:57
3^(8n+3) this means it will always be the third cycle.
3^1 = 3 3^2 = 9 3^3 = 27 3^4 = 81 and the cycle repeats.
Since we know 3^(8n+3) will have a units digit like the one in 3^3
Then we can do 7+2 = 9/5 = 1 4/5
Or similarly we can take 27+2 = 29/5 = 5 4/5
Answer choice E
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If n is a positive integer, what is the remainder when &nbs
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