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If n and m are positive integers, what is the remainder when 3^(4n+2)
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Updated on: 05 Jun 2019, 02:15
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If n and m are positive integers, what is the remainder when \(3^{(4n+2)} + m\) is divided by 10? (1) n = 2 (2) m = 1
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Originally posted by wshaffer on 12 Nov 2006, 13:38.
Last edited by Bunuel on 05 Jun 2019, 02:15, edited 1 time in total.
Renamed the topic and edited the question.




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Re: If n and m are positive integers, what is the remainder when 3^(4n+2)
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18 Nov 2010, 06:58
Geronimo wrote: If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10 ? (1) n=2 (2) m=1 When considering the remainder when a number is divided by 10, focus on the last digit of the number. That will be the remainder. e.g. 85/10, remainder 5. 39 divided by 10, remainder 9 etc... The last digit of powers of 3 have a cyclicity of 4. Look at the example below: 3^1 = 3 3^2 = 9 3^ 3 = 27 3^4 = 81 3^5 = 243 3^6 = 729 and so on.. notice the last digits 3, 9, 7, 1, 3, 9,.... thats the pattern they follow. So 3^(4n + 2) will end with 9. (If you are not comfortable with cyclicity, check out its theory. You can also check out this post where I have discussed cyclicity of some other numbers in detail: http://gmatclub.com/forum/cyclicityof103262.html#p803511) Now, we just need to know what m is. If m = 1, 3^(4n + 2) + m will end in 0. So remainder will be 0. Answer (B).
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Re: If n and m are positive integers, what is the remainder when 3^(4n+2)
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12 Nov 2006, 14:17
Given that n and m are positive integers, knowing n is useless because the expression (4n+2) will always give an answer which is 4 units apart. What we need to know is what m is for it will determine the remainder of the expression when divided by 10.
Notice that for any exponent of 3, the unit digit is repeating in cycles of 4:
3^1 = 3
3^2 = 9
3^3 = 2 7
3^4 = 8 1
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3^5 = 24 3
3^6 = 72 9
3^7 = 218 7
3^8 = 656 1
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Notice the cycle: 3971
Hence, knowing that m is 1, we know that the whole expression (4n+2)+m will give an exponent of 7, 11, 15, etc.
This ensures that the remainder will ALWAYS be 7 and B is sufficient.
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Re: If n and m are positive integers, what is the remainder when 3^(4n+2)
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06 Apr 2010, 20:31
mads wrote: If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10? 1. n=2 2. m=1 1: not enough as we do not know what is m hence impossible to answer remainder. 2: 3 has a cylicity of 4 i.e 3^1 = 3 3^2 = 9 3^3 = 27 3^4 = 81 3^5 = 243 hence cyclicity of 4 for any +ve value of n (1,2,3,.....) the unit digit of 3^(4n+2) will be 9 hence adding m=1 into it will give unit digit as 0 which is divisible by 10 hence sufficient to answer. Its B



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Re: If n and m are positive integers, what is the remainder when 3^(4n+2)
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06 Apr 2010, 21:48
mads wrote: If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10? 1. n=2 2. m=1 st 1) n=2, dont know about m. Not sufficient st 2) m=1 3^(4n+2) + m = 3^[2 * (2n+1)] + m = 9 ^(2n+1) + m 2n+1 is always odd . so the units digit of 9^(2n+1) is 9 and m is 1, so the remainder will be 0 B



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Re: If n and m are positive integers, what is the remainder when 3^(4n+2)
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20 Nov 2010, 18:38
VeritasPrepKarishma wrote: Geronimo wrote: If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10 ? (1) n=2 (2) m=1 When considering the remainder when a number is divided by 10, focus on the last digit of the number. That will be the remainder. e.g. 85/10, remainder 5. 39 divided by 10, remainder 9 etc... The last digit of powers of 3 have a cyclicity of 4. Look at the example below: 3^1 = 3 3^2 = 9 3^ 3 = 27 3^4 = 81 3^5 = 243 3^6 = 729 and so on.. notice the last digits 3, 9, 7, 1, 3, 9,.... thats the pattern they follow. So 3^(4n + 2) will end with 9. (If you are not comfortable with cyclicity, check out its theory. You can also check out this post where I have discussed cyclicity of some other numbers in detail: http://gmatclub.com/forum/cyclicityof103262.html#p803511) Now, we just need to know what m is. If m = 1, 3^(4n + 2) + m will end in 0. So remainder will be 0. Answer (B). if m=1 wouldn't it end in 7? I'm not sure how you got 0. 3^(4n + 2) + 1 = 3^(4n + 3) and based on the cyclicity pattern from above, it would be the next one over which will end in 7? maybe i'm not understanding this correctly, please correct me. thanks!



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Re: If n and m are positive integers, what is the remainder when 3^(4n+2)
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28 Nov 2019, 10:11
VeritasKarishma wrote: Geronimo wrote: If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10 ? (1) n=2 (2) m=1 When considering the remainder when a number is divided by 10, focus on the last digit of the number. That will be the remainder. e.g. 85/10, remainder 5. 39 divided by 10, remainder 9 etc... The last digit of powers of 3 have a cyclicity of 4. Look at the example below: 3^1 = 3 3^2 = 9 3^ 3 = 27 3^4 = 81 3^5 = 243 3^6 = 729 and so on.. notice the last digits 3, 9, 7, 1, 3, 9,.... thats the pattern they follow. So 3^(4n + 2) will end with 9. (If you are not comfortable with cyclicity, check out its theory. You can also check out this post where I have discussed cyclicity of some other numbers in detail: http://gmatclub.com/forum/cyclicityof103262.html#p803511) Now, we just need to know what m is. If m = 1, 3^(4n + 2) + m will end in 0. So remainder will be 0. Answer (B). I have a tough time with this not being C. Hear me out: We will not know what the exponent will be unless we know the value of N. Exponents change the value of the integer (obviously), and thus the final digit... 3 9 27 81 243 We will not know the remainder if we don't know what N is... what am I missing



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Re: If n and m are positive integers, what is the remainder when 3^(4n+2)
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01 Dec 2019, 20:31
esonrev wrote: VeritasKarishma wrote: Geronimo wrote: If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10 ? (1) n=2 (2) m=1 When considering the remainder when a number is divided by 10, focus on the last digit of the number. That will be the remainder. e.g. 85/10, remainder 5. 39 divided by 10, remainder 9 etc... The last digit of powers of 3 have a cyclicity of 4. Look at the example below: 3^1 = 3 3^2 = 9 3^ 3 = 27 3^4 = 81 3^5 = 243 3^6 = 729 and so on.. notice the last digits 3, 9, 7, 1, 3, 9,.... thats the pattern they follow. So 3^(4n + 2) will end with 9. (If you are not comfortable with cyclicity, check out its theory. You can also check out this post where I have discussed cyclicity of some other numbers in detail: http://gmatclub.com/forum/cyclicityof103262.html#p803511) Now, we just need to know what m is. If m = 1, 3^(4n + 2) + m will end in 0. So remainder will be 0. Answer (B). I have a tough time with this not being C. Hear me out: We will not know what the exponent will be unless we know the value of N. Exponents change the value of the integer (obviously), and thus the final digit... 3 9 27 81 243 We will not know the remainder if we don't know what N is... what am I missing The actual value of n is irrelevant. Say n = 1 3^(4n + 2) = 3^6  The units digit here will be 9 Say n = 2 3^(4n + 2) = 3^10  The units digit here will be 9 Say n = 3 3^(4n + 2) = 3^14  The units digit here will be 9 and so on... The exponent of 3 will always be of the form 4n + 2. So it will start a new cycle and end at the second term. So the units digit will always be 9. Even without the actual value of n, you know what you need to know  the units digit of 3^(4n + 2) is 9.
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If n and m are positive integers, what is the remainder when 3^(4n+2)
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01 Dec 2019, 21:20
wshaffer wrote: If n and m are positive integers, what is the remainder when \(3^{(4n+2)} + m\) is divided by 10?
(1) n = 2 (2) m = 1 To find remainder when the expression is divided by 10, calculate the unit digit of the given expression 3^(4n+2) + m = 3^4n * 3^2 + m Unit digit of 3^4n is 1 unit digit of 3^2 is 9 Require value of m to find unit digit of the expression (1) Insufficient (2) Sufficient B is correct.




If n and m are positive integers, what is the remainder when 3^(4n+2)
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01 Dec 2019, 21:20






