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16 Sep 2014, 01:51
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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) \(b = (c + 1)^3\)
(1) \(a = c^3\)
(2) \(b = (c + 1)^3\)
16 Sep 2014, 01:51
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Official Solution:
To determine the remainder after \(b - a\) is divided by 3, we need to express this difference as \(3n + r\), where \(n\) is an integer, and \(r\) is an integer equal to either 0, 1, or 2. The only bit of information that we have is that \(a\), \(b\), and \(c\) are positive integers.
(1) INSUFFICIENT. We do not have any information about \(b\), so we cannot determine the needed remainder.
(2) INSUFFICIENT. We do not have any information about \(a\), so we cannot determine the needed remainder.
(1) and (2) TOGETHER: SUFFICIENT. We can express \(b - a\) in terms of one variable (\(c\)):
\(b - a == (c + 1)^3 - c^3\)
\(== (c + 1)(c + 1)^2 - c^3\)
\(== (c + 1)(c^2 + 2c + 1) - c^3\)
\(== (c^3 + 2c^2 + c) + (c^2 + 2c + 1) - c^3\)
\(== c^3 + 3c^2 + 3c + 1 - c^3\)
\(== 3c^2 + 3c + 1\)
Look at the first two terms of this expression. \(3c^2\) and \(3c\) are both multiples of 3, since \(c\) is an integer. Their sum \(3c^2 + 3c\) is also a multiple of 3, equal to \(3(c^2 + c)\). The expression \(3c^2 + 3c + 1\) is equal to 1 more than a multiple of 3, and the remainder after division by 3 will always be 1.
Alternatively, we could plug numbers to find a pattern. If we list the first several cubes and take their differences, we see that these differences are always one more than a multiple of 3. This does not constitute proof, but we would suspect that the two statements together are sufficient.
\(1^3 = 1\)
\(2^3 = 8\)
\(3^3 = 27\)
\(4^3 = 64\)
\(5^3 = 125\)
\(5^3 - 4^3 = 61\), which leaves remainder 1 after division by 3
\(4^3 - 3^3 = 37\), which leaves remainder 1 after division by 3
\(3^3 - 2^3 = 19\), which leaves remainder 1 after division by 3
\(2^3 - 1^3 = 7\), which leaves remainder 1 after division by 3
Answer: C
To determine the remainder after \(b - a\) is divided by 3, we need to express this difference as \(3n + r\), where \(n\) is an integer, and \(r\) is an integer equal to either 0, 1, or 2. The only bit of information that we have is that \(a\), \(b\), and \(c\) are positive integers.
(1) INSUFFICIENT. We do not have any information about \(b\), so we cannot determine the needed remainder.
(2) INSUFFICIENT. We do not have any information about \(a\), so we cannot determine the needed remainder.
(1) and (2) TOGETHER: SUFFICIENT. We can express \(b - a\) in terms of one variable (\(c\)):
\(b - a == (c + 1)^3 - c^3\)
\(== (c + 1)(c + 1)^2 - c^3\)
\(== (c + 1)(c^2 + 2c + 1) - c^3\)
\(== (c^3 + 2c^2 + c) + (c^2 + 2c + 1) - c^3\)
\(== c^3 + 3c^2 + 3c + 1 - c^3\)
\(== 3c^2 + 3c + 1\)
Look at the first two terms of this expression. \(3c^2\) and \(3c\) are both multiples of 3, since \(c\) is an integer. Their sum \(3c^2 + 3c\) is also a multiple of 3, equal to \(3(c^2 + c)\). The expression \(3c^2 + 3c + 1\) is equal to 1 more than a multiple of 3, and the remainder after division by 3 will always be 1.
Alternatively, we could plug numbers to find a pattern. If we list the first several cubes and take their differences, we see that these differences are always one more than a multiple of 3. This does not constitute proof, but we would suspect that the two statements together are sufficient.
\(1^3 = 1\)
\(2^3 = 8\)
\(3^3 = 27\)
\(4^3 = 64\)
\(5^3 = 125\)
\(5^3 - 4^3 = 61\), which leaves remainder 1 after division by 3
\(4^3 - 3^3 = 37\), which leaves remainder 1 after division by 3
\(3^3 - 2^3 = 19\), which leaves remainder 1 after division by 3
\(2^3 - 1^3 = 7\), which leaves remainder 1 after division by 3
Answer: C
20 Feb 2016, 18:30
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Bunuel
If \(a\), \(b\), and \(c\) are positive integers, what is the remainder after \(b - a\) is divided by 3?
(1) \(a = c^3\)
(2) \(b = (c + 1)^3\)
(1) \(a = c^3\)
(2) \(b = (c + 1)^3\)
Can we tai the cube root of both \(a = c^3\) and \(b = (c + 1)^3\) to then plug into b - a?
* (c + 1) - c =1 --> 1/3 = 0 r 1
