# 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?

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.

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:$$\frac{876}{3}=222+\frac{210}{3}=222+70=292$$

• Mufasa, very beautiful way to divide. What happens when the denominator is 2 digits or the denominator is a single digit larger than the first digit in the numerator? 876/23 or 876/98 Sep 27, 2013 at 17:28
• $876/23$ would result in: 1. $23(0) + 8$, 2. $23(0) + 7$, 3. $23(0) + 6$. This would effectively leave you with a remainder of $876$ and so you would have to divide using the classical technique. Same goes for $876/98$. Sep 27, 2013 at 21:51