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# A DS question from GMATPrep

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Manager
Joined: 13 Mar 2007
Posts: 98

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A DS question from GMATPrep [#permalink]

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13 Oct 2007, 16:25
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

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!

Kudos [?]: 36 [0], given: 0

VP
Joined: 09 Jul 2007
Posts: 1098

Kudos [?]: 141 [0], given: 0

Location: London
Re: A DS question from GMATPrep [#permalink]

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13 Oct 2007, 16:40
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

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

Kudos [?]: 141 [0], given: 0

CEO
Joined: 29 Mar 2007
Posts: 2554

Kudos [?]: 516 [0], given: 0

Re: A DS question from GMATPrep [#permalink]

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13 Oct 2007, 23:15
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

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.

Kudos [?]: 516 [0], given: 0

Re: A DS question from GMATPrep   [#permalink] 13 Oct 2007, 23:15
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