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If n and m are positive integers, what is the remainder when 3^(4n+2) Updated on: 12 Aug 2022, 06:27
Originally posted by wshaffer on 12 Nov 2006, 14:38.
Last edited by Bunuel on 12 Aug 2022, 06:27, edited 2 times in total.
Renamed the topic and edited the question.
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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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If n and m are positive integers, what is the remainder when 3^(4n+2) 18 Nov 2010, 07:58
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Geronimo
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
Show more

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: https://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.
Answer (B).
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If n and m are positive integers, what is the remainder when 3^(4n+2) 06 Apr 2010, 21:31
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mads
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
Show more

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

:btw 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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If n and m are positive integers, what is the remainder when 3^(4n+2) 06 Apr 2010, 22:48
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mads
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
Show more
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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If n and m are positive integers, what is the remainder when 3^(4n+2) 20 Nov 2010, 19:38
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VeritasPrepKarishma
Geronimo
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: https://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.
Answer (B).
Show more

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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If n and m are positive integers, what is the remainder when 3^(4n+2) 28 Nov 2019, 11:11
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VeritasKarishma
Geronimo
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: https://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.
Answer (B).
Show more

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) 01 Dec 2019, 21:31
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VeritasKarishma
Geronimo
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: https://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.
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
Show more

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) 01 Dec 2019, 22:20
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wshaffer
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
Show more
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.
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If n and m are positive integers, what is the remainder when 3^(4n+2) 18 Mar 2020, 21:34
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VeritasKarishma
Geronimo
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: https://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.
Answer (B).
Show more

Quote:

Without doing all the math by trying different n values. Could you assume the exponent will always be the second one in the cyclicity because of the +2 in the exponent?
To better explain, if the exponent was (4n+1), could we assume that the units digit would be 3?
Show more

Absolutely! The "math" is to show you that cyclicality exists. Once you know it, you don't need to do it again.
3^(4n + 2) ends with a 9 for all non-negative integer values of n.
3^(4n + 1) ends with a 3 for all non-negative integer values of n.
etc
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If n and m are positive integers, what is the remainder when 3^(4n+2) 02 Aug 2020, 00:00
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wshaffer
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
Show more

Asked: If n and m are positive integers, what is the remainder when \(3^{(4n+2)} + m\) is divided by 10?

(1) n = 2
\(3^{(4n+2)} + m = 3^{10} + m \)
The remainder when \(3^{(4n+2)} + m\) is divided by 10 = Remainder when m is divided by 10
Since m is unknown
NOT SUFFICIENT

(2) m = 1
\(3^{(4n+2)} + m = 3^{(4n+2)} + 1 \)
The remainder when \(3^{(4n+2)} + m\) is divided by 10
Since \(3^4n = 81^n\), the remainder when \(3^{4n}\) when divided by 10 = 1
\(3^2 = 9\)
The remainder when \(3^{(4n+2)} \)when divided by 10 = 1 * 9 = 9
The remainder when \(3^{(4n+2)} + 1\) when divided by 10 = 0
SUFFICIENT

IMO B
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If n and m are positive integers, what is the remainder when 3^(4n+2) 21 Jul 2021, 12:55
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wshaffer
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
Show more

Stem Analysis:

3^4n+2 = (3^4n)(3^2)

So now we can break this down even further so that 9(3^4n)/10 + M/10. Now remember that bases to a power divided by the same denominator creates patterns. In this case any 3^4th power divided by 10 will leave some integer PLUS a remainder of 1. So 9(Q+1/10), in which Q is some integer. So now we have that 9Q + 9/10 + M/10. But we still don't know the value of M. If we have the value of M we can solve the question.

Statement 1
As analyzed in our stem the value of N doesn't help us, we need M. Not sufficient.

Statement 2
We have M, and so that means that the remainder must be zero because 9Q + 9/10 + 1/10 = 9Q + 1. Sufficient.
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If n and m are positive integers, what is the remainder when 3^(4n+2) 28 Oct 2021, 07:05
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VeritasKarishma
Geronimo
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: https://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.
Answer (B).
Show more

3^4n+2 ,
I looked at it this way: 4n will always be even, +2 will still keep it even
so thus 4n+2 is always even. the cyclicity of 3 is 3,9,7,1. So, I inferred that last digit would be 9 or 1
So, stat 2 wasnt sufficient

Now that Im not under time pressure, I tested numbers for n and saw that it always ends up being 9. But why is that? Why does it never end in 1?

Bunuel VeritasKarishma ScottTargetTestPrep @
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If n and m are positive integers, what is the remainder when 3^(4n+2) 28 Oct 2021, 20:53
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Chitra657
VeritasKarishma
Geronimo
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: https://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.
Answer (B).

3^4n+2 ,
I looked at it this way: 4n will always be even, +2 will still keep it even
so thus 4n+2 is always even. the cyclicity of 3 is 3,9,7,1. So, I inferred that last digit would be 9 or 1
So, stat 2 wasnt sufficient

Now that Im not under time pressure, I tested numbers for n and saw that it always ends up being 9. But why is that? Why does it never end in 1?

Bunuel VeritasKarishma ScottTargetTestPrep @
Show more

Chitra657

4n we know is an even number but we know something else about it too. That it is a multiple of 4.
Cyclicity of 3 is 4. So whenever the power is a multiple of 4, it will end in 1.
But here the power is 2 more than a multiple of 4 (it is 4n + 2) so the units digit of 3^(4n+2) will always be 9.
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