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by magical cook » Thu Nov 29, 2007 10:45 pm
What is the remainder when the positive integer n is divided by 3?
(1) The remainder when n is divided by 2 is 1.
(2) The remainder when n + 1 is divided by 3 is 2.


I dont have answer..so hope someone could help me with this. thanks
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by samirpandeyit62 » Thu Nov 29, 2007 11:00 pm
What is the remainder when the positive integer n is divided by 3?
(1) The remainder when n is divided by 2 is 1.

n is odd, some odd nos are evenly divisble by 3, some are not INSUFF

(2) The remainder when n + 1 is divided by 3 is 2.

n+1 = 3a +2

n =3a +1

so remainder is 1

SUFF

B
Regards
Samir
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by sujaysolanki » Thu Nov 29, 2007 11:03 pm
From 1 n = 1 3 5 7 9

n/3 varies hence insufficient

From 2

n = 1,4,7,10 Hence sufficient

Hence B
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by magical cook » Fri Nov 30, 2007 10:11 am
thank you!
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by bhumika.k.shah » Fri Jan 22, 2010 11:07 am
Is this a right approach though?

I aint denying that i used the same approach but in statement I we are getting many solutions and hence insufficient.
Likewise in Statement II ; however the answer is B

I did have a doubt in my mind but just chose B and went onto the next question...
sujaysolanki wrote:From 1 n = 1 3 5 7 9

n/3 varies hence insufficient

From 2

n = 1,4,7,10 Hence sufficient

Hence B
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by papgust » Sat Jan 23, 2010 6:49 am
Yeah that approach is right. You could either solve this by plugging in values (similar to that approach) or by setting up equations as samir has done.
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by Vena » Sat Nov 12, 2011 6:20 am
Consider stmt(1):The remainder when n is divided by 2 is 1.
n= 2K+1
where K is the quotient and 1 is the remainder
n/3 = (2K+1)/3
Take some examples:
K=2 -->(2K+1)/3 = 5/3, the remainder is 2
K=3 -->(2K+1)/3 = 7/3, the remainder is 1
There is no definite value for the remainder of n/3 --> stmt(1) is INSUFF

Consider stmt(2): The remainder when n + 1 is divided by 3 is 2.
(n+1)=3K+2
n= 3K+1
for every value of K, the remainder of (n+1)/3 is always 1
stmt(2) is SUFF
Choose B
Last edited by Vena on Sat Nov 12, 2011 6:22 pm, edited 1 time in total.
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by Luke.Doolittle » Sat Nov 12, 2011 7:33 am
bhumika.k.shah wrote:Is this a right approach though?

I aint denying that i used the same approach but in statement I we are getting many solutions and hence insufficient.
Likewise in Statement II ; however the answer is B
You could also look at (1) from a purely equational point of view (is equational a word?)

From the statement
n = 2k + 1
n = 2k + k + 1 - k
n = 3k + (1-k)

r = 1-k

Which is not a constant (varies with the quotient) and since you don't have a single value for r this is insufficient
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