For the product of two numbers to be even, one number must be even. For the product of two numbers to be odd, both numbers must be odd.

Remember:

$\displaystyle O \cdot E = E$

$\displaystyle O \cdot O = O$

$\displaystyle E \cdot E = E$

First solve by evaluating the ones digit.  Carry over a 1 from the $\displaystyle 10$ so that the value of the ones digit can be evaluated.

$\displaystyle 12-8=4$

After the carry over, the $\displaystyle 10$ becomes a $\displaystyle 9$.  This value will be used for the tens digit of the next term.  The hundreds digit will become zero.

Evaluate the tens digit.

$\displaystyle 9-1=8$

Combine the tens digit and the ones digit.

The final answer is $\displaystyle 84$.

1. First subtract the ones digits:

$\displaystyle 8-8=0$

So the ones digit will be a 0.

2. Next subtract the tens digit borrowing the 1 from the hundreds digit:

$\displaystyle 12-6=6$

So the tens digit will be a 6.

3. Add the tens and ones digits together:

$\displaystyle 60+0=60$

1. First subtract the ones digits borrowing a 1 from the tens digit:

$\displaystyle 14-8=6$

So the ones digit will be a 6.

2. Next subtract the tens digit borrowing the 1 from the hundreds digit:

$\displaystyle 16-8=8$

So the tens digit will be a 8.

3. Next add the hundreds, tens and ones digits together:

$\displaystyle 100+80+6=186$

The key here is for any integers $\displaystyle a$ and $\displaystyle b$, that means that no matter what you set $\displaystyle a$ and $\displaystyle b$ equal to you will get an odd number. An odd number is not divisible by 2, also it is an even number plus and odd number. The only expression that satisfies this is $\displaystyle 4b-2a+3$. $\displaystyle 4b$ will always be even, so will $\displaystyle -2a$, but $\displaystyle 3$ is always odd so the combination gives us an odd number, always.

For this problem, align and solve by adding the ones digit $\displaystyle (8+2)$, tens digit $\displaystyle (9+0+ carried\:1)$, and hundreds digit $\displaystyle (9+carried\:1)$. This also means that you have to add $\displaystyle 1$ to the $\displaystyle 1$ in the thousands place to get $\displaystyle 2$. 

$\displaystyle 998+1002=2000$

Rewrite the question into a mathematical expression.

$\displaystyle 12+42$

Add the ones digit. 

$\displaystyle 2+2=4$

Add the tens digit.

$\displaystyle 1+4=5$

Combine the tens digit and the ones digit. The answer is $\displaystyle 54$.

This problem can be solved using common factors.  

Rewrite $\displaystyle 80\div6$ using factors.  

$\displaystyle \frac{80}{6}=\frac{2 \cdot 40}{2 \cdot 3} = \frac{40}{3}$

Rewrite  $\displaystyle 1028\div1004$ by using common factors. Reduce to the simplest form.

$\displaystyle \frac{1028}{1004}=\frac{2\cdot 514}{2\cdot 502} = \frac{2\cdot 2\cdot257}{2\cdot 2\cdot251}=\frac{257}{251}$

Rewrite $\displaystyle 2032\div 234$ and use common factors to simplify.

$\displaystyle \frac{2032}{234}=\frac{2\cdot1016}{2\cdot 117}=\frac{1016}{117}$