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

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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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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.

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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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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
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]
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).

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]
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
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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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esonrev
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

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

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?

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

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

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

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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