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Interesting GMATFix Problem-13

by arora007 » Tue Sep 21, 2010 12:53 pm
When integer n is divided by integer p, the remainder is r. What is the sum, in terms of p and r, of all possible distinct remainders when n is divided by 3p?
A) 3r+3p
B) r+2p
C) 3r
D) 3p-r
E) 3r+p
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by fatalityish » Wed Sep 22, 2010 6:24 pm
Hey arora nice question.
the answer for this one is (A) 3p + 3r

I wish i could explain the solution how i did it. One way to solve this is by giving values to different parameters and checking.
I hope someone can come up with a better and faster solution.

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by limestone » Wed Sep 22, 2010 7:01 pm
IMO: A too
And this is my approach:
n/p --> remainder r, so n = p*a + r (with a is an integer)
then if n/3p then n = 3p*a/3 +r
If a is divisible to 3: a/3 is an integer, the remainder of n/3p is r
If a/3 to have the remainder is 1: n = 3p*(a-1)/3 + 3p*1/3 + r, the remainder is p + r (subtract 3p*a/3 to 3p*1/3 to create 3p*(a-1)/3; in which (a-1)/3 is an integer. I later add 3p*1/3 back to equalise the equation)
If a/3 to have the remainder is 2: n = 3p*(a-2)/3 + 3p*2/3 + r, the remainder is 2p + r
Sum of remainders in all cases: r + (p+r) + (2p+r) = 3r + 3p
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