# 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?

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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.

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

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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 – jessica Sep 27 '13 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 $876/98$. – Mufasa Sep 27 '13 at 21:51