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# M20-04

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:07
Expert's post
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Difficulty:

65% (hard)

Question Stats:

66% (01:47) correct 34% (02:17) wrong based on 148 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. $$A + C$$
[Reveal] Spoiler: OA

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16 Sep 2014, 00:07
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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, 18:54
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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, 09:41
I think this question is good and helpful.

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15 Apr 2015, 08:06
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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, 05:25
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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, 12:58
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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 bean 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, 17:25
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, 05:40
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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, 13:56
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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, 07:06
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, 03:01
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, 18:15
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+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, 15:41
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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Re: M20-04   [#permalink] 20 Nov 2017, 15:41
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# M20-04

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