Source: Manhattan Prep
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
The OA is C
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If a, b, and c are positive integers, what is the remainder
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BTGmoderatorLU
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Statement 1:BTGmoderatorLU wrote:Source: Manhattan Prep
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
No information about b.
INSUFFICIENT.
Statement 2:
No information about a.
INSUFFICIENT.
Statements combined:
Case 1: c=1, with the result that a=1³=1 and b=(1-2)³=-1
In this case, (a-b)/6 = [(1-(-1)]/6 = 2/6 = 0 R2
Case 2: c=3, with the result that a=3³=27 and b=(3-2)³=1
In this case, (a-b)/6 = (27-1)/6 = 26/6 = 4 R2
Case 3: c=4, with the result that a=4³=64 and b=(4-2)³=8
In this case, (a-b)/6 = (64-8)/6 = 56/6 = 9 R2
In every case, the remainder is 2.
SUFFICIENT.
The correct answer is C.
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Statement 1:BTGmoderatorLU wrote:Source: Manhattan Prep
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
No information about b.
INSUFFICIENT.
Statement 2:
No information about a.
INSUFFICIENT.
Statements combined:
Case 1: c=1, with the result that a=1³=1 and b=(1-2)³=-1
In this case, (a-b)/6 = [(1-(-1)]/6 = 2/6 = 0 R2
Case 2: c=3, with the result that a=3³=27 and b=(3-2)³=1
In this case, (a-b)/6 = (27-1)/6 = 26/6 = 4 R2
Case 3: c=4, with the result that a=4³=64 and b=(4-2)³=8
In this case, (a-b)/6 = (64-8)/6 = 56/6 = 9 R2
In every case, the remainder is 2.
SUFFICIENT.
The correct answer is C.
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$$\frac{\left(a-b\right)}{6}=??$$
Statement 1:
$$a=c^2\ or\ c=\sqrt[3]{a}$$
This only give the value of 'a' and 'c' while 'b' is unknown. Hence, statement 1 is INSUFFICIENT
Statement 2:
$$b=\left(c-2\right)^3\ $$
This only give the value of 'b' and 'c' while 'a' is unknown.
Hence, statement 2 is INSUFFICIENT.
Combining both statement together
$$a=c^3\ and\ b=\left(c-2\right)^3\ $$
$$a-b=c^3\ -\left(c-2\right)^3\ \ \ \ \ \ \ \left(difference\ of\ 2\ cubes\right)$$
$$a-b=\left(c-c-2\right)\left(c^2+c\left(c-2\right)+\left(c-2\right)^2\ \right)$$
$$a-b=\left(-2\right)\left(c^2+c^2-2c+c^2-4c+4\right)$$
$$a-b=-2c^2-2c^2+4c-2c^2+8c-8$$
$$a-b=-6c^2+12c-8$$
$$=-6\left(c^2+2c\right)-8$$
Dividing 8 by 6, the remainder = 2...
Therefore, statement 1 and 2 together are SUFFICIENT. Hence, OPTION C is correct
Statement 1:
$$a=c^2\ or\ c=\sqrt[3]{a}$$
This only give the value of 'a' and 'c' while 'b' is unknown. Hence, statement 1 is INSUFFICIENT
Statement 2:
$$b=\left(c-2\right)^3\ $$
This only give the value of 'b' and 'c' while 'a' is unknown.
Hence, statement 2 is INSUFFICIENT.
Combining both statement together
$$a=c^3\ and\ b=\left(c-2\right)^3\ $$
$$a-b=c^3\ -\left(c-2\right)^3\ \ \ \ \ \ \ \left(difference\ of\ 2\ cubes\right)$$
$$a-b=\left(c-c-2\right)\left(c^2+c\left(c-2\right)+\left(c-2\right)^2\ \right)$$
$$a-b=\left(-2\right)\left(c^2+c^2-2c+c^2-4c+4\right)$$
$$a-b=-2c^2-2c^2+4c-2c^2+8c-8$$
$$a-b=-6c^2+12c-8$$
$$=-6\left(c^2+2c\right)-8$$
Dividing 8 by 6, the remainder = 2...
Therefore, statement 1 and 2 together are SUFFICIENT. Hence, OPTION C is correct
