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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

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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

OA C

Source: Manhattan Prep

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BTGmoderatorDC wrote:
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

OA C

Source: Manhattan Prep
Note that each statement alone is not sufficient, so taking both the statements together.

We have to find out the remainder when a - b is divided by 6.

a - b = c^3 - (c - 2)^3 = c^3 - [c^3 - 3c^2*2 + 3c*2^2 - 2^3]

= c^3 - c^3 + 6c^2 - 24c + 8

= 6c^2 - 24c + 8

= (6c^2 - 24c + 6) + 2

We see that (6c^2 - 24c + 6) is divisible by 6, thus, the remainder = 2. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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BTGmoderatorDC wrote:
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
Statement 1:
No information about b.
INSUFFICIENT.

Statement 2:
No information about a.
INSUFFICIENT.

Statements combined:
Case 1: c=3, with the result that a=3³=27 and that b=(3-2)³=1
In this case, a-b = 27-1 = 26.
Dividing 26 by 6, we get:
26/6 = 4 R2

Case 2: c=4, with the result that a=4³=64 and that b=(4-2)³=8
In this case, a-b = 64-8 = 56.
Dividing 56 by 6, we get:
56/6 = 9 R2

Case 3: c=5, with the result that a=5³=125 and that b=(5-2)³=27
In this case, a-b = 125-27 = 98.
Dividing 98 by 6, we get:
98/6 = 16 R2

In every case, the remainder is 2.
SUFFICIENT.

The correct answer is C.

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BTGmoderatorDC wrote:
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
Source: Manhattan Prep
$$a,b,c\,\, \ge 1\,\,\,{\rm{ints}}\,\,\,\,\left( * \right)$$
$$a - b = 6Q + r\,\,,\,\,\,\left\{ \matrix{
\,Q,r\,\,\,{\mathop{\rm int}} {\rm{s}} \hfill \cr
\,0 \le r \le 5 \hfill \cr} \right.$$
$$? = r$$

$$\left( 1 \right)\,\,a = {c^3}\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,1,1} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 0 \hfill \cr
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {8,1,2} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 1 \hfill \cr} \right.$$

$$\left( 2 \right)\,\,b = {\left( {c - 2} \right)^3}\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,1,3} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 0 \hfill \cr
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {2,1,3} \right)\,\,\,\, \Rightarrow \,\,\,\,? = 1 \hfill \cr} \right.$$

$$\left( {1 + 2} \right)\,\,\,a - b = {c^3} - {\left( {c - 2} \right)^3} = \ldots = \underbrace {6{c^2} - 12c + 6}_{6Q\,\,,\,\,Q = {c^2} - 2c + 1} + 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 2$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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