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If the positive integer n is greater than 6, what is the

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If the positive integer n is greater than 6, what is the

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If the positive integer n is greater than 6, what is the remainder when n is divided by 6?

(1) When n is divided by 9, the remainder is 2.
(2) When n is divided by 4, the remainder is 1.

OA C

Source: Manhattan Prep

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BTGmoderatorDC wrote:
If the positive integer n is greater than 6, what is the remainder when n is divided by 6?

(1) When n is divided by 9, the remainder is 2.
(2) When n is divided by 4, the remainder is 1.
Here's a somewhat long approach (but it works!):

Target question: What is the remainder when n is divided by 6?

Given: Positive integer n is greater than 6

Statement 1: When n is divided by 9, the remainder is 2.
------ASIDE----------------
When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
------------------------------
So, from statement 1, we can conclude that some possible values of n are: 11, 20, 29, 38, 47, 56, 65. . . (since we're told n > 6, we can rule out n = 2) .
Let's test some possible values of n:
Case a: n = 11. When we divide 11 by 6, the remainder is 5
Case b: n = 20. When we divide 20 by 6, the remainder is 2
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: When n is divided by 4, the remainder is 1.
from statement 2, we can conclude that some possible values of n are: 9, 13, 17, 21, 25, 29, 33, 37, 41, . . . (since we're told n > 6, we can rule out n = 1 and n = 5) .
Let's test some possible values of n:
Case a: n = 9. When we divide 9 by 6, the remainder is 3
Case b: n = 13. When we divide 13 by 6, the remainder is 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that some possible values of n are: 11, 20, 29, 38, 47, 56, 65, . . .
Statement 2 tells us that some possible values of n are: 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69 . . .
From the above, we can see that some values of n that satisfy BOTH statements are: 29, 65, etc
We might also notice that if we keep adding 36 (the least common multiple of 9 and 4) to these possible n-values, we'll get more possible n-values such as 101, 137, 173, etc
Let's test some possible values of n:
Case a: n = 29. When we divide 29 by 6, the remainder is 5
Case b: n = 65. When we divide 65 by 6, the remainder is 5
Case c: n = 101. When we divide 101 by 6, the remainder is 5
Case d: n = 137. When we divide 137 by 6, the remainder is 5
So, it certainly looks like the remainder will always be 5.
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

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Brent Hanneson – Creator of GMATPrepNow.com
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BTGmoderatorDC wrote:
If the positive integer n is greater than 6, what is the remainder when n is divided by 6?

(1) When n is divided by 9, the remainder is 2.
(2) When n is divided by 4, the remainder is 1.
Source: Manhattan Prep
\[n \geqslant 7\,\,\,\operatorname{int} \,\,\,\,\left( * \right)\]
$$\left. \matrix{
n = 6Q + R\,\,\, \hfill \cr
Q\,\,\mathop \ge \limits^{\left( * \right)} \,\,1\,\,{\mathop{\rm int}} \hfill \cr} \right\}\,\,\,\,\,? = R\,\,\,\,\,\,\left( {0 \le R \le 5\,\,{\mathop{\rm int}} } \right)$$
\[\left( 1 \right)\,\,\,n = 9K + 2\,\,\,,\,\,\,K\,\mathop \geqslant \limits^{\left( * \right)} \,1\,\,\operatorname{int} \,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,K = 1\,\,\,\, \Rightarrow \,\,\,n = 11\,\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{5}}\,\,\,\,\,\left( {Q = 1} \right)\, \hfill \\
\,{\text{Take}}\,\,K = 2\,\,\,\, \Rightarrow \,\,\,n = 20\,\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{2}}\,\,\,\,\left( {Q = 3} \right) \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,\,n = 4J + 1\,\,\,,\,\,\,J\,\mathop \geqslant \limits^{\left( * \right)} \,2\,\,\operatorname{int} \,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,J = 2\,\,\,\, \Rightarrow \,\,\,n = 9\,\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{3}}\,\,\,\,\,\left( {Q = 1} \right)\, \hfill \\
\,{\text{Take}}\,\,J = 3\,\,\,\, \Rightarrow \,\,\,n = 13\,\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{1}}\,\,\,\,\left( {Q = 2} \right) \hfill \\
\end{gathered} \right.\]
$$\left( {1 + 2} \right)\,\,\,\,9K = n - 2\,\,\, = \,\,4J - 1\,\,\,\,\, \Rightarrow \,\,\,\,\,n - 2\,\,\,\,\left\{ \matrix{
\,{\rm{odd}}\,\,\,\, \Rightarrow \,\,\,n = {\rm{odd}}\,\,\,\, \Rightarrow \,\,\,R \in \left\{ {1,3,5} \right\} \hfill \cr
\,{\rm{multiple}}\,\,{\rm{of}}\,\,9,\,\,{\rm{hence}}\,\,{\rm{of}}\,\,3\,\,\,\, \Rightarrow \,\,\,n = 3M + 2 \hfill \cr} \right.\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,\,\,? = 5$$
$$\left( {**} \right)\,\,\,\,\left\{ \matrix{
\,R = 1\,\,\,\, \Rightarrow \,\,\,\,n = 6Q + 1\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 3\left( {2Q} \right) + 1\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 3M + 2\,\,\,\,\,{\rm{impossible}} \hfill \cr
\,R = 3\,\,\,\, \Rightarrow \,\,\,\,n = 6Q + 3\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 3\left( {2Q + 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,n = 3M + 2\,\,\,\,\,{\rm{impossible}} \hfill \cr} \right.\,\,\,\,\,\,\,\,$$

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

Regards,
Fabio.

_________________
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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