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If a , b , and c are positive integers, what is the remainder after b - a is divided by 3?

(1) a = c^3 (2) b = (c + 1)^3

If a , b , and c are positive integers, what is the remainder after b - a is divided by 3?

(1) a = c^3 --> no info about b. Not sufficient. (2) b = (c + 1)^3 --> no info about b. Not sufficient.

When taken together you can go with algebraic approach or plug-in method:

Algebraic approach:

(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 b-a --> b-a=(c + 1)^3-c^3=(c+1-c)(c^2+2c+1+c^2+c+c^2)=3c^2+3c+1=3(c^2+c)+1 --> remainder upon division this expression by 3 is 1. Sufficient.

Answer: C.

Plug-in method approach:

(1)+(2) try some numbers for a and b: a=c^3=1 --> b=(c+1)^3=8 and --> b-a=(c + 1)^3-c^3=7 --> remainder upon division 3 is 1; a=c^3=8 --> b=(c+1)^3=27 and --> b-a=(c + 1)^3-c^3=19 --> remainder upon division 3 is 1; a=c^3=27 --> b=(c+1)^3=64 and --> b-a=(c + 1)^3-c^3=37 --> remainder upon division 3 is 1; a=c^3=64 --> b=(c+1)^3=125 and --> b-a=(c + 1)^3-c^3=61 --> remainder upon division 3 is 1; ... It seem that there is some kind of pattern and we can safely assume that in all other cases remainder will also be 1. Sufficient.

Re: If a, b, and c are positive integers, what is the remainder [#permalink]
10 Oct 2012, 14:07

(A) & (B) ruled out as we do not get complete info about a & b together, because in question we are supposed to find remainder when (b-a) is divided by 3.

lets take both together;

Given: a=c^3;b=(c+1)^3 ) ( for a,b,c all positive integer) so lets check at three points; @c=1, a=1,b=8 ; @ c=2, a=8, b= 27, @ c= 3 , a=27,b=64

8-1/3 R->1 27-8/3 R->1 64-27/3 R->1

Hence C is the answer
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If a , b , and c are positive integers, what is the remainder after b - a is divided by 3?

(1) a = c^3 (2) b = (c + 1)^3

If a , b , and c are positive integers, what is the remainder after b - a is divided by 3?

(1) a = c^3 --> no info about b. Not sufficient. (2) b = (c + 1)^3 --> no info about b. Not sufficient.

When taken together you can go with algebraic approach or plug-in method:

Algebraic approach:

(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 b-a --> b-a=(c + 1)^3-c^3=(c+1-c)(c^2+2c+1+c^2+c+c^2)=3c^2+3c+1=3(c^2+c)+1 --> remainder upon division this expression by 3 is 1. Sufficient.

Answer: C.

Plug-in method approach:

(1)+(2) try some numbers for a and b: a=c^3=1 --> b=(c+1)^3=8 and --> b-a=(c + 1)^3-c^3=7 --> remainder upon division 3 is 1; a=c^3=8 --> b=(c+1)^3=27 and --> b-a=(c + 1)^3-c^3=19 --> remainder upon division 3 is 1; a=c^3=27 --> b=(c+1)^3=64 and --> b-a=(c + 1)^3-c^3=37 --> remainder upon division 3 is 1; a=c^3=64 --> b=(c+1)^3=125 and --> b-a=(c + 1)^3-c^3=61 --> remainder upon division 3 is 1; ... It seem that there is some kind of pattern and we can safely assume that in all other cases remainder will also be 1. Sufficient.

Answer: C.

Hope it helps.

Integers, when divided by 3 can leave a remainder of 0, 1, or 2. Integers cubed, when divided by 3, will leave the same remainders, because 0^3=0, 1^3=1, and 2^3=8=6+2.

Therefore, when subtracting cubes of two consecutive integers, the result will always leave a remainder of 1: the remainders repeat themselves cyclically 0,1,2,0,1,2,..., so 1-0=2-1=1 and \,\,0-2=-3+1. _________________

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