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M20-04
[#permalink]
16 Sep 2014, 01:07
16 Sep 2014, 01:07
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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 must be odd?
A. \(A + D\)
B. \(B + D\)
C. \(C + D\)
D. \(A + B\)
E. \(A + C\)
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A. \(A + D\)
B. \(B + D\)
C. \(C + D\)
D. \(A + B\)
E. \(A + C\)
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Re M20-04
[#permalink]
16 Sep 2014, 01:07
16 Sep 2014, 01:07
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Expert Reply
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 must be odd?
A. \(A + D\)
B. \(B + D\)
C. \(C + D\)
D. \(A + B\)
E. \(A + C\)
Subtract the first equation from the second:
\((D + B - A) - (A - C + B) = odd - even\);
\(C + D - 2A = odd\);
\(C + D = odd - 2A\);
\(C + D = odd - even\);
\(C + D = odd\).
Answer: C
_________________
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 must be odd?
A. \(A + D\)
B. \(B + D\)
C. \(C + D\)
D. \(A + B\)
E. \(A + C\)
Subtract the first equation from the second:
\((D + B - A) - (A - C + B) = odd - even\);
\(C + D - 2A = odd\);
\(C + D = odd - 2A\);
\(C + D = odd - even\);
\(C + D = odd\).
Answer: C
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General Discussion
M20-04
[#permalink]
20 Dec 2014, 19:54
20 Dec 2014, 19:54
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
(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
! |
This solution is NOT correct. Check the discussion below for more. |
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Re: M20-04
[#permalink]
15 Apr 2015, 09:06
15 Apr 2015, 09:06
22
Kudos
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.
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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Re: M20-04
[#permalink]
19 Jan 2016, 06:25
19 Jan 2016, 06:25
3
Kudos
1
Bookmarks
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
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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Re: M20-04
[#permalink]
20 Feb 2016, 13:58
20 Feb 2016, 13:58
14
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Bookmarks
I solved it in a different way:
add both expressions
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.
add both expressions
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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Joined: 31 Jul 2016
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Location: Singapore
Re: M20-04
[#permalink]
27 Nov 2016, 06:40
27 Nov 2016, 06:40
2
Kudos
1
Bookmarks
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
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
Re: M20-04
[#permalink]
06 May 2017, 14:56
06 May 2017, 14:56
1
Kudos
y4nkee wrote:
mvictor wrote:
I solved it in a different way:
add both expressions
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.
add both expressions
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
Re: M20-04
[#permalink]
08 Jun 2017, 08:06
08 Jun 2017, 08:06
1
Kudos
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.
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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Re: M20-04
[#permalink]
31 May 2018, 08:29
31 May 2018, 08:29
1
Kudos
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.
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.
Re: M20-04
[#permalink]
11 Jun 2018, 07:53
11 Jun 2018, 07:53
1
Kudos
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
-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
Re: M20-04
[#permalink]
02 Jul 2019, 16:51
02 Jul 2019, 16:51
1
Kudos
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.
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.
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Re: M20-04
[#permalink]
19 May 2020, 01:25
19 May 2020, 01:25
1
Bookmarks
A comprehensive solution :
To explain all options ;
E : even, O: Odd
A B C D
Options E E E O
O O E O
O E O E
E O O E
As we can see C, D are always of opposite parity.
To explain all options ;
E : even, O: Odd
A B C D
Options E E E O
O O E O
O E O E
E O O E
As we can see C, D are always of opposite parity.
Re: M20-04
[#permalink]
10 Sep 2020, 06:14
10 Sep 2020, 06:14
Hi,
we know that Odd - Even = Odd
we are given, D+B - A is odd
(Odd) (Even)
Which suggests D+B is odd, right? Not sure why this can't be the answer.
Any help appreciated, thanks in advance,
Ishita
we know that Odd - Even = Odd
we are given, D+B - A is odd
(Odd) (Even)
Which suggests D+B is odd, right? Not sure why this can't be the answer.
Any help appreciated, thanks in advance,
Ishita
Re: M20-04
[#permalink]
10 Sep 2020, 07:54
10 Sep 2020, 07:54
Expert Reply
ishitaz wrote:
Hi,
we know that Odd - Even = Odd
we are given, D+B - A is odd
(Odd) (Even)
Which suggests D+B is odd, right? Not sure why this can't be the answer.
Any help appreciated, thanks in advance,
Ishita
we know that Odd - Even = Odd
we are given, D+B - A is odd
(Odd) (Even)
Which suggests D+B is odd, right? Not sure why this can't be the answer.
Any help appreciated, thanks in advance,
Ishita
We are given that D + B - A = odd but it's possible that A is odd, B is odd, and D is odd. In this case D + B = even.
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Re: M20-04
[#permalink]
01 Feb 2022, 11:10
01 Feb 2022, 11:10
bartone89 wrote:
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
(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
How can we say, 2k and 2k+1, or (2) - (1) = 1. We don't know about the value of A, B, C or D.
Rearrangement does not seem correct using 2k or 2k + 1 values.
Can you explain if I am wrong or missing something?
Re: M20-04
[#permalink]
01 Feb 2022, 12:01
01 Feb 2022, 12:01
1
Kudos
Expert Reply
happypuppy wrote:
bartone89 wrote:
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
(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
How can we say, 2k and 2k+1, or (2) - (1) = 1. We don't know about the value of A, B, C or D.
Rearrangement does not seem correct using 2k or 2k + 1 values.
Can you explain if I am wrong or missing something?
No, you're not missing anything
That solution isn't correct -- it could become correct, if the first value was "2k" and the second value was "2m + 1", but using "k" both times isn't right. That solution is a bit inefficient though. We know D + B - A is odd and A - C + B is even. So if we subtract that second expression from the first one, we must get something odd, so D + B - A - (A - C + B) is odd, and simplifying, D + C - 2A is odd. Subtracting the even number 2A won't change evenness or oddness at all, so D + C is odd. 
Re: M20-04
[#permalink]
13 Mar 2023, 04:38
13 Mar 2023, 04:38
1
Kudos
I took the second equation and added the first equation to it.
the result is: 2B+D-C = odd
from this we can say that 2B is even so (D-C) has to be odd.
that means that either D = odd or C = even
or
D = even or C = odd
D and C cannot be odd at the same time or even at the same time.
is this logic good?
the result is: 2B+D-C = odd
from this we can say that 2B is even so (D-C) has to be odd.
that means that either D = odd or C = even
or
D = even or C = odd
D and C cannot be odd at the same time or even at the same time.
is this logic good?
Re: M20-04
[#permalink]
13 Mar 2023, 05:26
13 Mar 2023, 05:26
Expert Reply
joe123x wrote:
I took the second equation and added the first equation to it.
the result is: 2B+D-C = odd
from this we can say that 2B is even so (D-C) has to be odd.
that means that either D = odd or C = even
or
D = even or C = odd
D and C cannot be odd at the same time or even at the same time.
is this logic good?
the result is: 2B+D-C = odd
from this we can say that 2B is even so (D-C) has to be odd.
that means that either D = odd or C = even
or
D = even or C = odd
D and C cannot be odd at the same time or even at the same time.
is this logic good?
Yes. D - C = odd means that D + C = odd.
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Re: M20-04 [#permalink]
13 Mar 2023, 05:26
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