Remainders : Quant Question Archive [LOCKED]
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# Remainders

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Senior Manager
Joined: 20 Feb 2008
Posts: 296
Location: Bangalore, India
Schools: R1:Cornell, Yale, NYU. R2: Haas, MIT, Ross
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15 May 2008, 17:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Thanks alot for your help guys!
Director
Joined: 23 Sep 2007
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15 May 2008, 17:31
statement 1 is obviously not suff

statement 2 is not suff

together

the unit digit of 3^10 is 9
knowing the unit digit 9 and m=1
9+1 =10
10/10
remainder is 0
Intern
Joined: 05 May 2008
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15 May 2008, 18:20
Since we do not have a value for m, (1) is not sufficient.
For (2), the power of 3 is 4n+2.
For any value of n starting from 1,2.. the resulting number will end in a 9.

For n=1, 3$$^$$4n+2 =3$$^$$6=729.
3$$^$$4n+2 + 1 = 729 + 1 = 730.
730/10. Remainder is 0.

Same is the case for n=2,3 and so on. The number will always end in a 9 and tht when added 1, will result in a number directly divisible by 10 leaving a remainder 0.

Hence B.
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15 May 2008, 18:30
russell wrote:
Since we do not have a value for m, (1) is not sufficient.
For (2), the power of 3 is 4n+2.
For any value of n starting from 1,2.. the resulting number will end in a 9.

For n=1, 3$$^$$4n+2 =3$$^$$6=729.
3$$^$$4n+2 + 1 = 729 + 1 = 730.
730/10. Remainder is 0.

Same is the case for n=2,3 and so on. The number will always end in a 9 and tht when added 1, will result in a number directly divisible by 10 leaving a remainder 0.

Hence B.

I totally agree with this way of solving.
Senior Manager
Joined: 20 Feb 2008
Posts: 296
Location: Bangalore, India
Schools: R1:Cornell, Yale, NYU. R2: Haas, MIT, Ross
Followers: 4

Kudos [?]: 45 [0], given: 0

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15 May 2008, 18:35
Thanks alot, you are right, I made the mistake of assuming that 3^4n+2 is always going to be 3 raised to an even number which could result in the units digit of 9 or 1, hence a final value that ends in 0 or 2.
I should have worked it out!!
Re: Remainders   [#permalink] 15 May 2008, 18:35
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# Remainders

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