Pass the Evens

Similar reasoning: if $a+b$ is even and $b+c$ is even then their sum $a+2b+c$ is even. Subtracting an even $2b$ yields an even $a + c$

If $a+b$ is even and $b+c$ is even, then all three must be even or all three must be odd.

Here is why:

Suppose $a$ is odd. Then $a+b=$ even => $b$ is odd, and then $b+c$ =even => $c$ is odd. So all three are odd.

Suppose $a$ is even. Then $a+b=$ even => $b$ is even, and then $b+c$ =even => $c$ is even. So all three are even.

Either way, $\boxed{yes}$ $a+c$ is even.

For $a+b$ to be even, there are only two cases either both $a$ and $b$ are odd or both $a$ and $b$ are even.

Case 1: $a_{odd}+b_{odd}$

If both $a$ and $b$ are odd, for $b_{odd}+c$ to be even, $c$ must be odd. And $a_{odd}+c_{odd}$ is even .

Case 2: $a_{even}+b_{even}$

Similarly, if both $a$ and $b$ are even, for $b_{even}+c$ to be even, $c$ must be even. And $a_{even}+c_{even}$ is even .

In both cases $a+c$ is even, therefore the answer is $\boxed{\text{Yes}}$ .

Note that the sum of the three gives 2a + 2b + 2c, which is even, and subtracting (a+b) and (b+c) which are both even yields an even number, so (a+c) is even

b, being in both expressions, links the two and forces all terms to have same parity...so be that even or odd, any pair summed must be an even result.

An even number is the sum of two even numbers or the sum of two odd numbers,

Sum of the first two numbers = a +2b + c = 2b + (a +c)

2b is an even number

Therefore a + c is an even number.

One can easily prove that two integers $x$ and $y$ is of same parity if and only if $x$ $+$ $y$ is even.

" $a+b$ is even" implies $a$ is of same parity of $b$ , which in turn is of same parity of $c$ , as implied by " $b+c$ is even". Thus $a$ and $c$ must be of same parity.

So, $a+c$ is even.