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by edvhou812 » Tue May 10, 2011 8:32 pm
If n divided by 7 has a remainder of 2, what is the remainder when 3 times n is divided by 7?

A: 1
B: 2
C: 3
D: 5
E: 6

My apologies if this has been addressed. I couldn't find it in the forum. Please post answer through the use of formulas/equations if possible.
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by manpsingh87 » Tue May 10, 2011 10:05 pm
edvhou812 wrote:If n divided by 7 has a remainder of 2, what is the remainder when 3 times n is divided by 7?

A: 1
B: 2
C: 3
D: 5
E: 6

My apologies if this has been addressed. I couldn't find it in the forum. Please post answer through the use of formulas/equations if possible.
as n leaves 2 as a remainder when divided by 7,, hence it can be represented as..!!
n=7k+2; where k=1,2,3,4,5,,,,;
3n= 21k+6;
3n/7 = 21k+6/7 which will leave 6 as remainder hence E
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by GMATGuruNY » Wed May 11, 2011 3:58 am
edvhou812 wrote:If n divided by 7 has a remainder of 2, what is the remainder when 3 times n is divided by 7?

A: 1
B: 2
C: 3
D: 5
E: 6

My apologies if this has been addressed. I couldn't find it in the forum. Please post answer through the use of formulas/equations if possible.
A formula will make the problem seem harder than it is.
Just plug in a value for n that, when divided by 7, leaves a remainder of 2:

Let n=9.
9/7 = 1 R2, so n=9 works.
3n = 3*9 = 27.
27/7 = 3 R6.

The correct answer is E.
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by Brent@GMATPrepNow » Wed May 11, 2011 7:14 am
Mitch is right - it's often best to plug in numbers. Along that line, how do we find numbers that satisfy the given conditions?

There's a nice rule that goes like this:

If, when N is divided by D, the remainder is R, then the possible values of N include: R, R+D, R+2D, R+3D,. . .

Aside: the rule assumes that N, D and R are non-negative integers

I mention this because students often overlook the first option (R)

So, if we're told that n divided by 7 has a remainder of 2, then we know that the possible values of n include 2, 9, 16, 23, etc
2 is a possible value of n because if we divide 2 by 7 we get 0 with remainder 2

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by edvhou812 » Wed May 11, 2011 6:54 pm
Thank you all for your answers. Each of them were helpful.
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