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

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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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

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
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3^5 = 243
3^6 = 729
3^7 = 2187
3^8 = 6561
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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.

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

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!

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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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.

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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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

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.

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 @

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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