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 \(\,\,0-2=-3+1.\)
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