If A, B and C are integers, is A + B + C even?

Given that A, B, and C are three positive integers.
We are to determine if A+B+C = an even number.
There two ways that this is possible. If all three numbers are even or if only one is even.

Statement 1: A - C - B is even
This is sufficient since statement 1 is only possible if one of the three integers is even or all the integers are even. And we know that A+B+C can only be even if all three numbers or if one of the integers is even. Statement 1 implies one of the three numbers is even hence sufficient.

Statement 2: (A-C)/B is odd
This is also sufficient. This because if both A and C are even, A-C is even, and to get an odd number, then B must also be even, since an even number can be an odd multiple and dividing by an appropriate even number will lead to an odd number. The converse, on the other hand, is not true. An odd number cannot be an even multiple. Secondly, When either A or C is odd, A-C will be odd, hence the only way to get (A-C)/B to be odd is for B to be odd, implying we will get one even number which is sufficient to conclude that A+B+C is even.
When both A and C are odd, then A-C is even, and as stated earlier, we can only get (A-C)/B to be odd if B is odd. So we end up with one even number and two odd numbers and A+B+B is even.

The answer is therefore D.

Statement 1:
Let A-C-B = 0, with the result that A = B+C.
If A=B+C=odd, then A + (B+C) = ODD + ODD = EVEN.
If A=B+C=even, then A + (B+C) = EVEN + EVEN = EVEN.
In each case, the answer to the question stem is YES.
SUFFICIENT.

Statement 2:
Let (A - C)/B = 1, with the result that A-C = B and A = B+C.
Same equation as that yielded by Statement 1.
Since Statement 1 is sufficient, Statement 2 must also be SUFFICIENT.

odd+-even=odd
odd+-odd=even
even+-even=even

1. o+o+o=e+o=o
2. e+e+o=e+o=o
3. e+e+e=e+e=e
4. o+o+e=e+e=e

(1) A - C - B is even sufic

a-c-b=even; case 3, and 4 fit, both give an even sum.

(2) (A - C)/B is odd sufic

e/e=[odd, even, proper fraction, undefined]
o/o=[odd, proper fraction]
e/o=[even, proper fraction]
o/e=[undefined, proper fraction]

a-c/b=odd;
b=o: a-c=odd=even-odd; a,b,c=e+o+o=even.
b=e: a-c=even=even-even or odd-odd; a,b,c=e+e+e=even, and e+o+o=even.

Ans (D)

If A, B and C are integers, is A + B + C even?

(Statement1) A - C - B is even
Let's say that A - C - B=2k.
A=2k+B+C --> A + B + C= 2k+B+C+ B+C= 2k+2B+2C= 2(k+B+C) -EVEN (Always YES)

Sufficient

(Statement2) \(\frac{(A - C)}{B}\) is odd
Let's say that \(\frac{(A - C)}{B}= 2k+1\)

A= B*(2k+1)+ C

--> A + B + C =B*(2k+1)+ C+ B+C= B*(2k+1+1) + 2C= B*(2k+2)+2C= 2B*(k+1)+ 2C -EVEN (Always YES)
Sufficient

The answer is D.

a+b+c ; even ; e+e+e or o+o+e
#1
A - C - B is even
e-e-e or e-o-o
sufficient
#2
(A - C)/B is odd
e-e/e = Yes
(e-o)/o= yes
(o-o)/e = yes

we are getting values in e+e+e or o+o+e
sufficient
IMO D


If A, B and C are integers, is A + B + C even?

(1) A - C - B is even

(2) (A - C)/B is odd