Let's prove that by contradiction.

Suppose that we can express ${\color{#3D99F6}a} + {\color{#20A900}b}$ , ${\color{#3D99F6}a} + {\color{#D61F06}c}$ and ${\color{#20A900}b} + {\color{#D61F06}c}$ as integers. That is, for some integers $k_1$ , $k_2$ and $k_3$ , $\begin{array}{rl} {\color{#3D99F6}a} + {\color{#20A900}b} &= 2k_1 + 1\\ {\color{#3D99F6}a} + {\color{#D61F06}c} &= 2k_2 + 1\\ {\color{#20A900}b} + {\color{#D61F06}c} &= 2k_3 + 1 \end{array}$ Adding these expressions altogether, we have $2\left({\color{#3D99F6}a} + {\color{#20A900}b} + {\color{#D61F06}c}\right) = 2\left(k_1 + k_2 + k_3\right) + 3$ Since $LHS$ (left-hand side) is even for any choice of integers ${\color{#3D99F6}a}$ , ${\color{#20A900}b}$ and ${\color{#D61F06}c}$ , whereas $RHS$ (right-hand side) is odd, $\boxed{\text{all 3 sums can't be odd}}$ .

Given that,

• $a ,$ $b,$ and $c$ are integers.

• It is possible for all three sums to be even.

We can get ''even'' by $Even +$ $Even$ $+$ $Even =$ $Even$ so if $(a + b ) + (a + c) + (b + c)$ are even then there sum cannot be odd.

If $(a + b ), (a + c), (b + c)$ are all odd numbers then there sum should be also odd, let's check it.

$(a + b ) + (a + c) + (b + c) = 2a + 2b + 2c$

$(a + b ) + (a + c) + (b + c) = 2(a + b + c)$

We can see that $2(a + b + c)$ is even therefore, $\large\color{#3D99F6} \boxed{it~ is ~ not ~ possible}.$

Given that a, b, c are three integers and asked to check a+b , a+c, b+c all are odd numbers.

Sum of two integers is odd, if and only if one number must be odd and another must be even.

It is not possible to have three odd and three even numbers since we will get the result as even numbers.

If we have two odd and one even number, let say a=even, b=odd, c=odd then b+c is even.

So it is not possible that a+b , a+c, b+c all are odd numbers.

It can be solved using contradiction

At first taking that (a+b), (b+c), (c+a) all are odd.

Now it is obvious that, odd+odd+odd=odd

Again (a+b)+(b+c)+(c+a)=(a+b+c)×2 which is even.

But it is found in previous case that the sum should be odd.

Therefore, contradiction!

So it is not possible.

Notice that an even number plus an even number is even, and an odd number plus an odd number is also even. A positive integer can only either be odd or even. (For negative integers, "even" means congruent to 0 modulo 2, "odd" means congruent to 1 modulo 2.) Since we have 3 integers, we see that at least two of them have to be the same parity by the Pigeonhole Principle, and their sum is guaranteed to be even. Thus, the pairwise sum of the three integers cannot be all odd. $\blacksquare$

Easy. An odd number being (always) an even number + an odd number, you can't have all three sums odd. Proof, by possible settings:

all even numbers - obvious. not working.

one odd number - the two other numbers are even, generating an even sum.

two odd numbers - the two odd numbers added will give an even number.

all odd numbers - see previous setting.

Voila

If (a+b) is odd and (a+c) is odd then b and c are either both odd (if a is even) or both even (if a is odd). In either case, (b+c) will be even.

First, let's just assume that all three of those quantities are odd. Since we assume they are all odd, we can add two of those numbers together and we will get an even number (odd + odd = even).

So, $(a+b) + (a+c)$ must be even, but by simply rearranging the terms we see that the expression is equivalent to $(a+a) + (b+c)$ .

We can easily show $a+a$ must be even, but we also know that $b+c$ is odd by our assumption....but an even plus an odd is odd. So, by rearranging the terms we get a different result; this is a contradiction . Then it must be true that all three of those quantities cannot be odd so the answer is no .

By the Dirichlet's box principle, at least 2 of the numbers must have same parity. And adding those two numbers will always give an even number.Hence contradiction!!! So all 3 sums can't be odd.

Let's solve by exhaustion. If all 3 of a, b, c are even then a + b, a +c, and b + c are even. If 2 are even: let's say a and b are even and c is odd (without loss of generality). Then a + b is even. If only 1 is even: let's say that a is even and b and c are odd (without loss of generality). Then b + c is even. If all 3 of a, b, c are odd, then a + b, a + c, and b + c are even. So it is not possible that all 3 of a + b, a + c, and b + c are odd.

Maybe less mathematical: If a is odd, both b and c must be even to yield odd sums. If a is even, both b and c must be odd to yield odd sums. Then b + c must be the sum of two odd or two even numbers; in either case, it is even.

there can be three possibilities:

possibility-1.............all the numbers are even.so the answer will always be even..............[even+even=always even]

possibility-2.............all the three number are odd.suppose, they are 1, 3 & 7.

                 so, 1+3=4,  3+7=10. ,  7+1=8...............................[every time they are even]

possibility-3..............among 3 there are even and odd integers.suppose, they are 1,2 & 3

           so, 1+2=3[odd],  2+3=5[odd],   1+3=4[even]

so, it is not possible for all of the sum to be odd

Take a=c=1 and b=2 which satisfy the above condition.

in a, b and c, there must either be two evens or two odds. All pairs are summed, including the matching ones, which will be even.

It's simple no two odd number and two even numbers sum can be odd So its"NO"

Simply put: There are 6 integers, and since 6 is an even number, 6 multiplied by any number (odd or even) it will always produce an even number result.

I like to solve as follows (similar to other solutions):

There are 8 possible even/odd (e/o) permutations on integers a, b, c. Enumerating the permutations and sums:

a b c a+b a+c b+c

e e e e e e

e e o e o o

e o o o o e

o o o e e e

o o e e o o

o e e o o e

e o e o e o

o e o o e o

Therefore all three sums will never be odd.

To result in an odd number, you need to sum a pair number to an odd number. But if we have the sum that envolves all the three numbers, you will have a sum of either odd with odd or pair with pair. So no, it is not possible.