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dukenukem
If a, b and c are integers, is abc divisible by 4?

(1) a + b + 2c is even.
(2) a + 2b + c is odd.


I shall say, be flexible.

statement 1 says: either a & b are both even OR a & b are both odd.
a=3, b=3 c=2 Answer is NO
a=3, b=3, c=4 Answer is YES

Not sufficient

statement 2 says: a+c is odd.
so either of a or c is even
Still insufficient.

On combining both of these info, we are still not sure if any of these is an even multiple of 2.
Not sufficient.
+1 E
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Thank you! And also thanks for those links Bunuel!

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And apologies, spoiler: You guys are right! E is the right answer. wow so many ways to get there :( (sad face only cause I didn't realize there were so many methods when solving myself!)

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Can't we solve this problem by knowing that a, b, or c could be zero? Zero is an integer and it is even.
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Hi FWU,

YES, you COULD use 0 in any of your TESTs and come up with a quick "YES" answer. You still have to be thorough enough with your work to find one of the various "NO" answers that also exists though, so that you can get to the correct answer.

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dukenukem
If a, b and c are integers, is abc divisible by 4?

(1) a + b + 2c is even.
(2) a + 2b + c is odd.
\(a,b,c\,\,{\rm{ints}}\,\,\,\,\left( * \right)\)

\({{abc} \over 4}\,\,\mathop = \limits^? \,\,{\mathop{\rm int}}\)

\(\left( {1 + 2} \right)\,\,\,\,\)

\(\left( 1 \right)\,\,\,{{a + b + 2c} \over 2} = {\mathop{\rm int}} \,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\,{{a + b} \over 2} = {\mathop{\rm int}} \,\,\,\,\)

\(\left( 2 \right)\,\,\,{{a + 2b + c} \over 2} \ne {\mathop{\rm int}} \,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\,{{a + c} \over 2} \ne {\mathop{\rm int}}\)


\(\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {0,0,1} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,1,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.\,\,\,\)


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

Regards,
Fabio.
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dukenukem
If a, b and c are integers, is abc divisible by 4?

(1) a + b + 2c is even.
(2) a + 2b + c is odd.

Hi Everyone,

Not sure if this has been posted before, but here is my question: I'm using the manhattan gmat advanced quant book (just because I tried doing a GMAT practice test and found that I wasn't doing as well as I hoped on the quant - 70th percentile) and I found the suggestion to use scenario charts very interesting. However, when I try using them, I can't get my time to go below 3:30s to 4 mins which I know is too long to be working on any one problem.
I'm sure people have had success with the tool, which is why MGMAT recommends it; however, does anyone have any idea on how to use the chart more efficiently?

As an example:
I'm using the chart for try-it problem 4-7 in the MGMAT advanced quant book:
If a,b and c are integers, is abc divisible by 4?

(1) a + b + 2c is even.
(2) a + 2b + c is odd.


I have 5 columns: a, b, c, 4th column: to ensure a+b+2c is even and 5th column to check if abc is divisible by 4. I try different values for a, b and c and etc till I find whether abc is always divisible by 4 or not.

MGMAT suggests to look at a, b and c as O/E numbers. I find this would not have been intuitive for me (since the question asks "is it divisible by 4," I would automatically start by trying out numbers.)

Anyways, if anyone has some tips or something that you think I should do different, please feel free to share. Any help will be appreciated!

Thanks!!
Official Explanation:
Evaluate the possible odd/even combinations of a, b, and c using Scenario Charts to determine the answer to the question.
(1) INSUFFICIENT: 2c must be even because c is an integer. This statement implies that a + b = Even, which occurs when a and b have the same odd/even parity. (There is no constraint on c.) An efficient Scenario Chart might look like this: Please see the attachment (Statement 1 image)

(2) INSUFFICIENT: 2b must be even because b is an integer. Thus, this statement implies that a + c = Odd, which occurs when a and c have opposite odd/even parity. (There is no constraint on b.) -->Please see the attachment (Statement 2 image)

(1) AND (2) INSUFFICIENT: From (1) you know that a and b have the same odd/even parity, while from (2) you know that a and c have opposite odd/even parity.--> Please see the attachment (Statement 1&2 image)
Even with these constraints, you do not have a definitive answer to the question. The correct answer is (E).
Notice that in evaluating statements (1) and (2) together, you would not need to write a completely new Scenario Chart. You can reuse the work from statement (1) and statement (2) to determine the answer (for instance, by circling the scenarios in one chart that also appear in the other chart). Just be careful as you do this! Know what case you're considering.
Attachments

Statement 1 & 2 image.PNG
Statement 1 & 2 image.PNG [ 50.55 KiB | Viewed 3415 times ]

Statement 2 image.PNG
Statement 2 image.PNG [ 50.94 KiB | Viewed 3413 times ]

Statement-1 image.PNG
Statement-1 image.PNG [ 57.7 KiB | Viewed 3398 times ]

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fskilnik
dukenukem
If a, b and c are integers, is abc divisible by 4?

(1) a + b + 2c is even.
(2) a + 2b + c is odd.
\(a,b,c\,\,{\rm{ints}}\,\,\,\,\left( * \right)\)

\({{abc} \over 4}\,\,\mathop = \limits^? \,\,{\mathop{\rm int}}\)

\(\left( {1 + 2} \right)\,\,\,\,\)

\(\left( 1 \right)\,\,\,{{a + b + 2c} \over 2} = {\mathop{\rm int}} \,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\,{{a + b} \over 2} = {\mathop{\rm int}} \,\,\,\,\)

\(\left( 2 \right)\,\,\,{{a + 2b + c} \over 2} \ne {\mathop{\rm int}} \,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\,{{a + c} \over 2} \ne {\mathop{\rm int}}\)


\(\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {0,0,1} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,1,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.\,\,\,\)


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

Regards,
Fabio.
fskilnik
Sir, could you tell me how did you get statement 1 and Statement 2 insufficient with notations and rationale method?
Thanks__
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