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# If n and m are positive integers, what is the remainder when 3^(4n+2)

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If n and m are positive integers, what is the remainder when 3^(4n+2)  [#permalink]

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Updated on: 05 Jun 2019, 03:15
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69% (01:24) correct 31% (01:38) wrong based on 526 sessions

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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

Originally posted by wshaffer on 12 Nov 2006, 14:38.
Last edited by Bunuel on 05 Jun 2019, 03:15, 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)  [#permalink]

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18 Nov 2010, 07:58
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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/cyclicity-of-103262.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.
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Re: If n and m are positive integers, what is the remainder when 3^(4n+2)  [#permalink]

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06 Apr 2010, 21:31
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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

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)  [#permalink]

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12 Nov 2006, 15:17
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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 = 27
3^4 = 81
____________________________
3^5 = 243
3^6 = 729
3^7 = 2187
3^8 = 6561
____________________________
Notice the cycle: 3-9-7-1
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.

do not provide the answer before it is answered, nobody may attempt it. Give it a maximum of 2 days and provide the answer thereafter.
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Re: If n and m are positive integers, what is the remainder when 3^(4n+2)  [#permalink]

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06 Apr 2010, 22:48
1
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)  [#permalink]

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01 Dec 2019, 21:31
1
esonrev 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/cyclicity-of-103262.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.

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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Re: If n and m are positive integers, what is the remainder when 3^(4n+2)  [#permalink]

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20 Nov 2010, 19:38
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/cyclicity-of-103262.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.

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)  [#permalink]

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28 Nov 2019, 11:11
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/cyclicity-of-103262.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.

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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If n and m are positive integers, what is the remainder when 3^(4n+2)  [#permalink]

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01 Dec 2019, 22:20
1
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)   [#permalink] 01 Dec 2019, 22:20
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