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# If a, b, and c are positive integers, what is the remainder

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If a, b, and c are positive integers, what is the remainder  [#permalink]

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Updated on: 25 Jun 2014, 01:59
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73% (01:02) correct 27% (01:28) wrong based on 387 sessions

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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

Originally posted by Smita04 on 03 Feb 2012, 21:45.
Last edited by Bunuel on 25 Jun 2014, 01:59, edited 2 times in total.
Edited the question and added the OA
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Re: If a, b, and c are positive integers, what is the remainder  [#permalink]

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04 Feb 2012, 03:55
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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.

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.

Hope it helps.
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Re: If a, b, and c are positive integers, what is the remainder  [#permalink]

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10 Oct 2012, 15:07
(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

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Re: If a, b, and c are positive integers, what is the remainder  [#permalink]

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10 Oct 2012, 16:01
1
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.

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.

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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Re: If a, b, and c are positive integers, what is the remainder  [#permalink]

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04 Oct 2017, 07:14
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Re: If a, b, and c are positive integers, what is the remainder &nbs [#permalink] 04 Oct 2017, 07:14
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