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IMO C

O*O=O
Two entities have to be odd for the product to be odd.
For 1, c=O, a+b should be O, therefore a can be O or E & accordingly b can be O or E. Not Sufficient.

For 2, b=O, a+c should be O, therefor a can be O or E & accordingly c can be O or E. Not Sufficient.

1 & 2, we know that for both the statements to hold true, b and c are O. Therefore, a = E. Sufficient.
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chetan2u
Is a an even integer, if a,b and c are positive integers?

(1) (a+b)*c is odd
(2) (a+c)*b is odd

New question
source-self



S1.

a and b are the unique relevant terms here.
a and b can be either even or odd.


INSUFF.





S2.

a and c are the unique relevant terms here.
a and c can be either even or odd.


INSUFF.






S1.2.

Case 1. a is even; b and c are odd.
Case 2. a is odd; b and c are even.


INSUFF.






AC: E




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chetan2u
Is a an even integer, if a,b and c are positive integers?

(1) (a+b)*c is odd
(2) (a+c)*b is odd

New question
source-self



S1.

a and b are the unique relevant terms here.
a and b can be either even or odd.


INSUFF.





S2.

a and c are the unique relevant terms here.
a and c can be either even or odd.


INSUFF.






S1.2.

Case 1. a is even; b and c are odd.
Case 2. a is odd; b and c are even.


INSUFF.






AC: E




-


Hi

The second case that you have taken - 'a is odd, b & c are even' - i dont think that case is possible. Because if b is even, then (a+c)*b will become even and the second statement will no longer hold true. So I think answer should be C only. Can you recheck once again?
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Correct me if I am wrong but...

From statement 2 we can deduce that C is ODD.
If we combine it with statement 1 we have 2 possible cases (I assume that A=2)

(2+1)^3 = 3^3 = 27
(2-5)^3 = -3^3 = -27

In the case 1 A B and C are positive
In the case 2 we have B negative!

Hence i picked E! Why the answer is C?
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MvArrow
Correct me if I am wrong but...

From statement 2 we can deduce that C is ODD.
If we combine it with statement 1 we have 2 possible cases (I assume that A=2)

(2+1)^3 = 3^3 = 27
(2-5)^3 = -3^3 = -27

In the case 1 A B and C are positive
In the case 2 we have B negative!

Hence i picked E! Why the answer is C?

Hi

Few things to note here:

1) We CANNOT deduce from statement 2 that 'c' is odd. Product of (a+c) and b is given to be odd. This tells us that 'b' is odd for sure, and one out of a/c is odd while other out of a/c is even. First statement tells us that 'c' is odd.

2) When we combine the two statements, we can conclude since both b & c are odd, a has to be even in order to have both conditions to be true.

3) you have yourself taken the case of b as negative, it is nowhere given in the question. We are specifically given that a, b, c are positive integers. So we have to take only those values. We cannot take b as -5 here.
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chetan2u
Is a an even integer, if a,b and c are positive integers?

(1) (a+b)*c is odd
(2) (a+c)*b is odd

New question
source-self

Hi chetan2u & amanvermagmat

Can you help me to clarify my doubt.

I got the answer as C by combining both statements via one approach. However, when I do it with the below approach I am getting different answer.

\(a * c + b * c = odd\)
\(a * b + c * b = odd\)

Subtract the above two equations, we get

\(a (c-b) = even\)... As c & b are already odd as per the two statements
We can have a = odd or even.

Not sure where i am going wrong.
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chetan2u
Is a an even integer, if a,b and c are positive integers?

(1) (a+b)*c is odd
(2) (a+c)*b is odd

New question
source-self

Hi chetan2u & amanvermagmat

Can you help me to clarify my doubt.

I got the answer as C by combining both statements via one approach. However, when I do it with the below approach I am getting different answer.

\(a * c + b * c = odd\)
\(a * b + c * b = odd\)

Subtract the above two equations, we get

\(a (c-b) = even\)... As c & b are already odd as per the two statements
We can have a = odd or even.

Not sure where i am going wrong.

Hi Rahul

I think your approach is not wrong, its just that here this approach is incomplete.
Because this approach does not tell us what 'a' will be - odd or even?

However, as 'b' and 'c' are already odd as per statements, then instead of subtracting the two equations, why not use this for any one equation only: say first equation.

a*c + b*c = Odd, now b & c are both odd so b*c is odd. So a*c = Odd - Odd = even.

Product of a & c is even, but since c is odd, 'a' definitely has to be even.

(we can do the same thing for second equation also and we will get the same result)


What you have done is not wrong, but not very useful towards the required answer. Consider this:- say we are given
x+y = 5
2x + y = 9 and say we have to find x.
If I subtract first equation from second, I will straight away get x.
But instead if I add the two equations, I will get 3x + 2y = 14. What I have now done is not wrong, but it doesnt help me get x. I will have to do something more with this, or do something else to get x.
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chetan2u
Is a an even integer, if a,b and c are positive integers?

(1) (a+b)*c is odd
(2) (a+c)*b is odd

New question
source-self
Good question

1. c is definitely odd
a,b can be
1,2
2,1

2. b is definitely odd
a,c can be
1,2
2,1

1+2
If b and c are both odd then only RED option can prevail. So, a=even C
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