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Re: If a, b, and c are positive integers, what is the remainder [#permalink]
04 Feb 2012, 02:55

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Smita04 wrote:

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 _________________

" Make more efforts " Press Kudos if you liked my post

Re: If a, b, and c are positive integers, what is the remainder [#permalink]
10 Oct 2012, 15:01

1

This post received KUDOS

Bunuel wrote:

Smita04 wrote:

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. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If a, b, and c are positive integers, what is the remainder [#permalink]
24 Jun 2014, 23:52

Bunuel wrote:

Smita04 wrote:

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.

Hi Bunuel, Would it be possible for you to provide OA to the question and adjust the difficulty level, so that we guys can actually have some appreciation for the question?

Thanks! _________________

"GMAT conquered. Let's talk about applications..."

Persistence is the key. And I'm gonna make through...

Re: If a, b, and c are positive integers, what is the remainder [#permalink]
25 Jun 2014, 01:00

1

This post received KUDOS

Expert's post

neo656 wrote:

Bunuel wrote:

Smita04 wrote:

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.

Hi Bunuel, Would it be possible for you to provide OA to the question and adjust the difficulty level, so that we guys can actually have some appreciation for the question?