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10 Oct 2012, 14:56
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
68% (01:49) correct
32%
(02:02)
wrong
based on 559
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If a, b, and c are positive integers, what is the remainder when a – b is divided by 6?
(1) a = c^3
(2) b = (c – 2)^3
(1) a = c^3
(2) b = (c – 2)^3
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
10 Oct 2012, 15:15
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thevenus
If a, b, and c are positive integers, what is the remainder when a – b is divided by 6?
(1) a = c^3. No info about b. Not sufficient.
(2) b = (c – 2)^3. No info about b. 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 \(a-b\) --> \(a-b=c^3-(c-2)^3=(c-c+2)(c^2+c^2-2c+c^2-4c+4)=2(3c^2-6c+4)=6c^2-12c+8=6(c^2-2c+1)+2\) --> remainder upon division this expression by 6 is 2. Sufficient.
Answer: C.
Hope it helps.
10 Oct 2012, 15:33
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thevenus
Obviously, neither (1), nor (2) alone is sufficient.
Integers, when divided by 6 leave remainders 0, 1, 2, 3, 4, or 5.
When raised to the third power, the remainders stay the same 0, 1, 2, 3, 4, or 5:
\(6^3\)- remainder 0
\(1^3\) - remainder 1
\(2^3=8=6+2\) - remainder 2
\(3^3=27=24+3\) - remainder 3
\(4^3=64=60+4\) - remainder 4
\(5^3=125=120+5\) - remainder 5
Therefore, when subtracting the cubes of two integers two units apart, the difference will leave a remainder of 2 when divided by 6:
\(2-0=3-1=4-2=5-2\) and \(0-4=1-5=-4=-6+2.\)
Remainders repeat themselves cyclically \(0,1,2,3,4,5,0,1,2,3,4,5...\)
Answer C.
