Another Approach to Long division I need your help with a lifelong problem I have always had. There are two things in life I hate most, 1) weddings and 2) Long division. For the life of me I hate division. I am horrible at it. I can multiple large numbers in my head but I can’t divide if  my life depended on it. When the division operation is 2 digits over 1 or 2 digit, it’s not a major issue. But when I divide anything 3+ over a 2 digit number, it becomes an issue.
I am trying to find a way t divide that appeals to my abilities to multiply. I am trying to find another way to approach a division problem than the typical long division approach.
For example: 876/2. Simple enough Ans: 438. This isn’t a problem, Finding the new approach is.
I am trying to break this division number into single digits. 
$$8=2(4)+0$$
$$7=2(3)+1$$
$$6=2(3)+0$$
My problem is how does this approach get me to 438? The numbers in the parenthesis get me to 433 with a summed remainder of 1. If I carry the one and put it on the first digit of my number, 3, I get 434.
In typical long division, the remainder is carried over. So 
$$8=2(4)+0$$
$$7=2(3)+1 $$(the remainder 1 is carried down)
$$16=2(8)+0$$
Is there a way to do long division by doing the division individually on each number and then summing the remainder?
 A: What you did as:1. $8=2(4)+0$2. $7=2(3)+1$3. $6=2(3)+0$
Any remainder in step 1 is in the hundreds column
Any remainder in step 2 is in the tens column
Any remainder in step 3 is in the units column
So, in your case, you had a remainder of $1$ in step 2, which means your remainder is effectively $1*10=10$.
So, to complete your division, you would say:$$\frac{876}{2}=433+\frac{10}{2}$$
Now the problem you will end up with if you continue your strategy to divide $10$ by $2$ is as follows:1. $1=2(0)+1$ (i.e. remainder $=1*10=10$)2. $0=2(0)+0$
This means you get left with:$$\frac{10}{2}=0+\frac{10}{2}$$
i.e. you will recurse infinetly without reaching a result.
But, if you modify your strategy to stop at this point and do the final division in the classical manner, then you would get:$$\frac{876}{2}=433+\frac{10}{2}=433+5=438$$
As another example, consider 876/3:1. $8=3(2)+2$2. $7=3(2)+1$3. $6=3(2)+0$
This gives you:$$\frac{876}{3}=222+\frac{210}{3}=222+70=292$$
A: Integer division (in decimal form) usually amounts to iteratively finding part of the quotient, subtracting it off, then continuing.
This is what you've done: you've determined that you can write
$$876 = 433 \cdot 2 + 10 $$
When you get the remainder of $10$, you can't just "carry" it or "add" it: you have to divide it by $2$ too.
Or, find something else to do with it. e.g. change the three "digits" from 8,7,6 to 8,6,16 and try dividing again.
