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# A DS question from GMATPrep

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Joined: 13 Mar 2007
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A DS question from GMATPrep [#permalink]

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13 Oct 2007, 15:25
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If you know how to solve this problem and you can explain your answer, please let me know.

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

The correct answer is: B

Here what I did:

According to a formula that found online, whenever we have x divided by y we can write it as: x = y + remainder.

Now we have: 3^(4n+2) + m = 10 + remainder, thus we need to k now n and m in order to find remainder. However, according to the solution, we only need to know m. So, I am getting confuse about this question. Please help me if you know how to solve this problem. Thanks!
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Re: A DS question from GMATPrep [#permalink]

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13 Oct 2007, 15:40
GMAT_700 wrote:
If you know how to solve this problem and you can explain your answer, please let me know.

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

The correct answer is: B

Here what I did:

According to a formula that found online, whenever we have x divided by y we can write it as: x = y + remainder.

Now we have: 3^(4n+2) + m = 10 + remainder, thus we need to k now n and m in order to find remainder. However, according to the solution, we only need to know m. So, I am getting confuse about this question. Please help me if you know how to solve this problem. Thanks!

Gmat does not throw questions that can not be solved by following easy steps.

3^(4n+2)+m

1. n=2, 3^10+m we cannot say anything without knowing the value, it is smth. (...7+m)/10 m can be anything and can yield any remainder.

2. m=1, 3^(4n+2)+1

as we know n is a positive integer. thus put any number from 1 to million

notice
3^1=3
3^2=9
3^3=27
3^4=81
3^5=243
3^6=...9 when n=1
3^7=..7
3^8=..1
3^9=..3
3^10=..9 when n=2

3^14=..9
when n=3

and so on

so there is no remainder when 3^(4n+2)+1 as it is smth ...0/10

B
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Re: A DS question from GMATPrep [#permalink]

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13 Oct 2007, 22:15
Ravshonbek wrote:
GMAT_700 wrote:
If you know how to solve this problem and you can explain your answer, please let me know.

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

The correct answer is: B

Here what I did:

According to a formula that found online, whenever we have x divided by y we can write it as: x = y + remainder.

Now we have: 3^(4n+2) + m = 10 + remainder, thus we need to k now n and m in order to find remainder. However, according to the solution, we only need to know m. So, I am getting confuse about this question. Please help me if you know how to solve this problem. Thanks!

Gmat does not throw questions that can not be solved by following easy steps.

3^(4n+2)+m

1. n=2, 3^10+m we cannot say anything without knowing the value, it is smth. (...7+m)/10 m can be anything and can yield any remainder.

2. m=1, 3^(4n+2)+1

as we know n is a positive integer. thus put any number from 1 to million

notice
3^1=3
3^2=9
3^3=27
3^4=81
3^5=243
3^6=...9 when n=1
3^7=..7
3^8=..1
3^9=..3
3^10=..9 when n=2

3^14=..9
when n=3

and so on

so there is no remainder when 3^(4n+2)+1 as it is smth ...0/10

B

Well said.
There is a thread on this problem already b/c I remember writing on it.
Re: A DS question from GMATPrep   [#permalink] 13 Oct 2007, 22:15
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# A DS question from GMATPrep

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