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# M20-04

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Math Expert
Joined: 02 Sep 2009
Posts: 58398

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16 Sep 2014, 01:07
3
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Difficulty:

65% (hard)

Question Stats:

67% (02:24) correct 33% (02:37) wrong based on 129 sessions

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If $$A$$, $$B$$, $$C$$, and $$D$$ are integers such that $$A - C + B$$ is even and $$D + B - A$$ is odd, which of the following expressions is always odd?

A. $$A + D$$
B. $$B + D$$
C. $$C + D$$
D. $$A + B$$
E. $$A + C$$

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16 Sep 2014, 01:07
3
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Official Solution:

If $$A$$, $$B$$, $$C$$, and $$D$$ are integers such that $$A - C + B$$ is even and $$D + B - A$$ is odd, which of the following expressions is always odd?

A. $$A + D$$
B. $$B + D$$
C. $$C + D$$
D. $$A + B$$
E. $$A + C$$

Rewrite $$A - C + B$$ as $$(A + B) - C$$ and $$D + B - A$$ as $$D + (B - A)$$. If $$A + B$$ is even, $$B - A$$ is even too. If $$A + B$$ is odd, $$B - A$$ is odd too. Now it follows from the stem that:
• if $$A + B$$ is even, $$C$$ is even and $$D$$ is odd
• if $$A + B$$ is odd, $$C$$ is odd and $$D$$ is even

Thus, $$C + D$$ is always odd.

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20 Dec 2014, 19:54
25
1
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I think I may have found a much more simple methodology to answer this question:

(1) A - C + B = even; thus, A - C + B = 2k (where k is an integer)

(2) D + B - A = odd; thus, D + B - A = 2k + 1 (where k is an integer)

rearrange (2): D + B - A - 1 = 2k

Sub: rearranged (2) into (1): A - C + B = D + B - A - 1 --> 2A + 1 = C + D

thus, per my method above, if C + D = 2A + 1 (where A is an integer), C + D is ALWAYS ODD
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08 Jan 2015, 10:41
I think this question is good and helpful.
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15 Apr 2015, 09:06
14
bartone89 I believe your method is wrong because it assumes the values of the two equations are consecutive. ie 2k and 2k+1. I hope I havent misunderstood you solution.

Anyway, I think I may have found a better method.

The two equations are A-C+B and D+B-A, even and odd respectively.

On subtracting the two, you get, A-C+B-D-B+A = 2A-C-D (which is odd, since even - odd = odd)

2A-(C+D) is odd. Which means C+D is odd since we know that 2A has to be even.

C+D is one of the options.
Hope this makes sense.
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19 Jan 2016, 06:25
1
1
There is an easier way this question can be solved.
Lets say A-C+B = 2n (even)
and after rearranging the second equation, -A+D+B=2m+1 (odd)

Adding the two equations we get, 2B+D-C=2(m+n)+1

From this, 2B is even and 2(m+n) is even, hence we conclude,
D-C=odd

When difference of two integers is odd, there sum will also be odd;

Hence D+C = odd
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20 Feb 2016, 13:58
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I solved it in a different way:
A-C+B+D+B-A=odd.
ok, rewrite everything as:
2B+D-C=odd
2B is always even, it must mean only 1 thing: D-C is always odd. if D-C is odd, then D+C is always odd.
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26 Apr 2016, 18:25
mvictor wrote:
I solved it in a different way:
A-C+B+D+B-A=odd.
ok, rewrite everything as:
2B+D-C=odd
2B is always even, it must bean only 1 thing: D-C is always odd. if D-C is odd, then D+C is always odd.

Very good solution mvictor!
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Re: M20-04 I solved it with a different approach! Just Sharing  [#permalink]

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27 Nov 2016, 06:40
2
1
Add or Substract the two equations-

A-C+B=Even
D+B-A=Odd
=2A+C-D (Even-Odd is always Odd) and we know 2A is always even

Therefore C-D should be Odd and thus also C+D
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06 May 2017, 14:56
1
y4nkee wrote:
mvictor wrote:
I solved it in a different way:
A-C+B+D+B-A=odd.
ok, rewrite everything as:
2B+D-C=odd
2B is always even, it must bean only 1 thing: D-C is always odd. if D-C is odd, then D+C is always odd.

Very good solution mvictor!

I agree very good solution. But it still requires timing luck or incredible number sense. I solved it through plugging in values. My method take no number sense needed and does not require too much logic.

A-C+B is even. Just plug in some easy numbers so that is true. A = 1 C = 2 B = 3 means that 1-2+3 = 2 which is even.
D+B-A is odd. Re-use the same variables from before. D + 3 - 1 is D + 2. Set D to an easy odd number which is 5. So D = 5.
Now you have values for A through D or (A,B,C,D) = (1,3,2,5). Work through the answers.

A. A+D = 1 + 5 = 6
B. B+D = 3 + 5 = 8
C. C+D = 2 + 5 = 7. This is odd. No need to do the rest, the answer is C.
D. A+B
E. A+C
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08 Jun 2017, 08:06
Given A-C+B =Even and D+B-A=Odd
Subtract 1) and 2)
2A-C-D=Even-odd=Odd=> 2A=Odd- (C+D)
Left side of the equation is even hence C+D has to be Odd.
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25 Jun 2017, 04:01
Given A-C+B=Even-----1
& D+B-A = Odd---------2
Add 1 & 2 ---even + odd = odd
hence C-D= 2B - Odd= Odd
if C-D is odd, then C+D is also Odd. (eg 4-3=1, 4+3=7)
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18 Sep 2017, 19:15
+A-c+B= even
D+B-A = odd Now equation 1 - equation2 which gives 2A-(D+C)=odd
2A is always even so (D+C) is always odd
Hence C
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20 Nov 2017, 16:41
joondez wrote:
y4nkee wrote:
mvictor wrote:
I solved it in a different way:
A-C+B+D+B-A=odd.
ok, rewrite everything as:
2B+D-C=odd
2B is always even, it must bean only 1 thing: D-C is always odd. if D-C is odd, then D+C is always odd.

Very good solution mvictor!

I agree very good solution. But it still requires timing luck or incredible number sense. I solved it through plugging in values. My method take no number sense needed and does not require too much logic.

A-C+B is even. Just plug in some easy numbers so that is true. A = 1 C = 2 B = 3 means that 1-2+3 = 2 which is even.
D+B-A is odd. Re-use the same variables from before. D + 3 - 1 is D + 2. Set D to an easy odd number which is 5. So D = 5.
Now you have values for A through D or (A,B,C,D) = (1,3,2,5). Work through the answers.

A. A+D = 1 + 5 = 6
B. B+D = 3 + 5 = 8
C. C+D = 2 + 5 = 7. This is odd. No need to do the rest, the answer is C.
D. A+B
E. A+C

the thing is - when presented with such expressions, you have to either add or subtract one from another.
the thing is, you have 3 variables in each expression, therefore, you might lose some extra time to test all the possible options...
my approach is very direct and fast though...and I always have the same approach to solving similar questions.
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31 May 2018, 08:29
The basic Rule to solve this question is if
A+B = even then A-B will also be even and vice versa
Same way if A+B is odd the A-B will also be odd and vice versa
Now
A-C+B = Even ---->1
D+B-A = Odd ---->2
adding 1 & 2 we have
D-C +2B = Odd
D-C = Odd- 2B
D-C = Odd

hence D+C is odd.
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09 Jun 2018, 01:02
How to approach question in this way.
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11 Jun 2018, 07:53
1
A - C + B = even
-A+D +B = ODD
---------------------
D - C + 2B = ODD --> 2B is always odd.

so D-C = ODD
Therefore, D+C = ODD

Please let me know if the approach is correct and can be followed

Thanks & Warm Regards,
Richin
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23 Oct 2018, 19:48
Bunuel is there a way i can find similar problems? i tried searching for similar prblms but all i could find are gmatclub test problems.... please suggest a way... and i request a way to find similar topics ...youre too kind to provide the links but i cannot trouble every single time...
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23 Oct 2018, 20:38
Bunuel is there a way i can find similar problems? i tried searching for similar prblms but all i could find are gmatclub test problems.... please suggest a way... and i request a way to find similar topics ...youre too kind to provide the links but i cannot trouble every single time...

Check number properties tag here: https://gmatclub.com/forum/search.php?view=search_tags
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02 Jul 2019, 16:51
I solved by making a small chart to list possibilities:

If A-C+B = E E E
-A+D+B = E O E

If A-C+B = O E O
-A+D+B = O O O

By checking the answers against this, you can see that only C+D will be E+O = Odd number both ways.
Re: M20-04   [#permalink] 02 Jul 2019, 16:51
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# M20-04

Moderators: chetan2u, Bunuel