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Re: Dividing cubicle [#permalink]
29 Oct 2012, 04:37

Expert's post

If P, Q, and R are positive integers, what is the remainder when R – Q is divided by 3?

(1) R = P^3. No info about Q. Not sufficient. (2) Q = (P – 2)^3. No info about R. Not sufficient.

(1)+(2) Important tip: x^3-y^3 can be factored as follows:x^3-y^3=(x-y)(x^2+xy+y^2). Apply this factoring to R-Q --> R-Q=P^3-(P-2)^3=(P-P+2)(P^2+P^2-2P+P^2-4P+4)=2(3P^2-6P+4)=6c^2-12c+8=6(c^2-2c+1)+2 --> remainder upon division this expression by 3 is 2. Sufficient.

Re: If P, Q, and R are positive integers, what is the remainder [#permalink]
10 Aug 2013, 07:39

Hi folks,

Both are insufficient without any doubts checking for option (C).

P=3 , P-2 =1 27-1 = 26/3 =2

P=4 , P-2 = 2 64-8 = 56/3 =2

P=5 , P-2 = 3 125-27 , Rem=2

P=6, P-2 = 4 Rem =2

Hence (C)
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Re: If P, Q, and R are positive integers, what is the remainder
[#permalink]
10 Aug 2013, 07:39