Long division has always been a weakness of mine and some how I've gotten through school and sixth form without it, but i'd like to learn it, it's just that I have a problem with intuition. So I know this may seem silly to a lot of people but I'd really appreciate it if you could guide me through the process of division and provide me with some explanation of what is happening at each step. I know the general process is divide, multiply, subtract, but what is exactly going on here? Could someone explain this to me with regards to the above question if it isnt too much trouble, or a better type of question that would illustrate division intuitively. Thank you.

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    $\begingroup$ its not easy to coach such basics here try Khan academy videos I am sorry it's just if you didn't understand it by now i think it's hard to pick up from answer's here watch a video on Khan academy $\endgroup$ – MRK Sep 15 '13 at 23:54

Here, for reference, is the long division, in its usual form on the left and in an expanded form on the right:

             4026                          4026  
            -----                         -----  
          6)24158                       6)24158  
            24                            24000  
            ---                           -----  
              1                             158  
              0                               0  
              --                            ---  
              15                            158  
              12                            120  
              ---                           ---  
               38                            38  
               36                            36  
               --                            --  
                2                             2

I begin by approximating $24158$ as $24000$. When I divide $6$ into $24$ and get $4$, what I’m really doing is dividing $24000$ by $6$ to get $4000$ and writing down just the $4$, in a position that implies that there are $3$ more digits to follow. The multiplication $4\cdot6=24$ is standing in for $6\cdot4000=24000$, and the subtraction $24-24$ is standing in for $24158-24000$. This is shown explicitly in the version on the right. After the subtraction, we see that $158$ out of the original $24158$ remains unaccounted for: $4000$ $6$’s are only enough to give a total of $24000$. Again, this is spelled out fully in the calculation on the right.

In the standard calculation on the left we write down only those digits that we need at each step. Subtracting a $0$ does nothing, so there’s no need to write down the $000$ in $24000$. I also don’t need the entire remainder of $158$ at this point. When I bring down the $1$, I am simply approximating $158$ by a single hundred; this does not contain any positive multiple of $6$ hundreds, so I write a $0$ in the quotient in the position indicating the number of hundreds. I then improve the approximation by bringing down the $5$ and thinking of $158$ as $150$, or $15$ tens. There are $2$ full $6$’s in $15$, so $150$ contains $2$ tens of $6$’s, or $20$ $6$’s; I indicate this by writing a $2$ in the quotient in the position indicating the number of tens. That accounts for $2$ tens of $6$’s, or $6\cdot20=120$; this product is explicit on the right, while on the left I omit the $0$.

To say the same thing in slightly different words, we first determine how many thousands of $6$’s can be accommodated in $24158$, namely, $4$; those $4$ thousands of $6$’s account for $24000$ of the $24158$, leaving the $6$’s in $158$ to be counted. We then determine how many hundreds of $6$’s can be accommodated in this remainder of $158$, and the answer is $0$: a single hundred of $6$’s would already be $600$, which is far too big. Thus, we still have $158$ to account for. How many tens of $6$’s will it accommodate? Just $2$, since $3$ tens of $6$’s is $180$. That takes care of another $120$, leaving only $38$, which accommodates $6$ single $6$’s. And that takes care of another $36$, leaving only $2$ unaccounted for. We can either report the result as $4026$ with a remainder of $2$, meaning that $24158=6\cdot4026+2$, or continue the process by seeing how many tenths of $6$’s can be squeezed into the remainder of $2$, which is, after all, the same as $20$ tenths. $3$ tenths of $6$’s come to $0.3\cdot6=1.8$, which ‘fits’ inside the remainder of $2$, while $4$ tenths of $6$’s comes to $2.4$, which is too large, so the next digit (i.e., the first digit to the right of the decimal point) would be $3$.

You can also think of division as a kind of repeated subtraction: how many times can you subtract $6$ from $24158$ without dropping below $0$, and how much is left at the end? In this case you can subtract $6$ a total of $4026$ times, and you’ll have $2$ left over at the end. But instead of actually subtracting one $6$ at a time, we do it in handy bundles. Since $$10,000\le24,158<100,000\;,$$ we start with bundles of $10,000$ $6$’s. Each bundle contains $60,000$ units, and $60,000>24,158$, so we can’t subtract even one of these bundles, and we drop to bundles of $1000$ $6$’s. Each of them contains $6000$ units, so we can subtract $4$ of them without dropping below $0$. That leaves $158$. We can’t subtract any bundles of a hundred $6$’s, but we can subtract $2$ bundles of ten $6$’s, leaving $38$. Finally, we can subtract $6$ $6$’s from $38$, leaving $2$ leftover units. Altogether we subtracted $4$ bundles of a thousand $6$’s each, $0$ bundles of a hundred $6$’s each, $2$ of ten $6$’s each, and $6$ single $6$’s, so we subtracted a total of $$4\cdot1000+0\cdot100+2\cdot10+6\cdot1=4026$ $6$’s.

  • $\begingroup$ How would I apply this if the divisor is double figures or more? $\endgroup$ – seeker Sep 16 '13 at 14:43
  • $\begingroup$ @Assad: You’re welcome! If the divisor is larger, the idea remains exactly the same; it’s just harder to tell by eye how many times it will ‘fit’. If you’re dividing $314$ into $456342$, for instance, you’ll start by seeing how many bundles of a thousand $314$’s there are; there’s one, and subtracting $314000$ leaves you with $142342$. Now you have to determine how many bundles of a hundred $314$’s will fit into that; it turns out to be $4$, which account for another $125600$, leaving $16742$. That accommodates $5$ bundles of ten $314$’s, leaving $1042$, which accommodates $3$ $314$’s, ... $\endgroup$ – Brian M. Scott Sep 16 '13 at 16:28
  • $\begingroup$ ... with $100$ left over as a remainder. (Whew!) $\endgroup$ – Brian M. Scott Sep 16 '13 at 16:28
  • $\begingroup$ Sorry, I just saw this now. Thank you for your insight Brian, it's cleared up a lot of misunderstandings, i'm indebted to you, thank you again for your carefully written and intelligent responses. $\endgroup$ – seeker Sep 16 '13 at 22:50
  • $\begingroup$ @Assad: You’re very welcome. $\endgroup$ – Brian M. Scott Sep 16 '13 at 22:56

First start checking digit by digit starting from the left

Is 2 bigger than or equal to 6? No. So we move one.

$\left \lfloor\frac{2}{6}\right \rfloor = 0$. But because $0$ at the beginning of the number don't matter we skip adding them to the answer.

Is 24 bigger than or equal to 6? Yes. It is.

Now we start the so called "DMS (Divide, Multiply, Subtract) loop".

$\left \lfloor\frac{24}{6}\right \rfloor = 4$ So $4$ is our first digit of the answer. Now we multiply $4 \cdot 6 = 24$ and we subtract that from 24 and we'll end up with $0$. Now we add all the remaining digits to this number, so this leaves us with 158. So we repeat again.

Is 1 bigger than or equal to 6? No. So we move one.

$\left \lfloor\frac{1}{6}\right \rfloor = 0$. But this time we add zero, because we already have a digit in the answer.

Is 15 bigger than or equal to 6? Yes. It is.

Now again we apply DMS.

$\left \lfloor\frac{15}{6}\right \rfloor = 2$. So $2$ is our third digit no we multiply $2 \cdot 6 = 12$ and subtract that from $15$. This leaves us with $3$. Now we add the remaining digits, actually a digit in this case. So now we have $38$

Is 38 bigger than or equal to 6? Yes. It is.

$\left \lfloor\frac{38}{6}\right \rfloor = 6$. Now we add 6 to the answer and subtract $6 \cdot 6 = 36$ from $38$. This leaves us with $2$. Beacuse this is our last digit we stop (Note taht if we wanted to get a decimal we need to add $0$'s to $2$, but we have to add a comma to the answer).

So $\frac {24158}{6} = 4026$ and remainder $2$.

  • $\begingroup$ Thank you so much! It was the thing with the placement of the 0's that confused me! $\endgroup$ – seeker Sep 16 '13 at 12:22

A possible 'intuitive guidepost', is to always have an estimate of the answer before you begin. One thing to always remember is that when using the $\text{Base-}10$ system, you are an 'instant expert' at multiplying and dividing by $10$. So, take the 'lazy approach' and get an estimate.

Now $6,000$ goes into $24,158$, but $60,000$ is no good. So the answer is between $1,000$ and $10,000$. Since $6 \times 4 = 24$, you know the answer is over $4,000$. You might also intuitively see that the final answer is 'real close to' $4,000$.

After getting this estimate you can work on the problem using long division. But now you are so close, just continue 'carving out' the answer:

$\frac {24,158}{6,000} = 4 + \frac{158}{6,000} \text{ iff }$
$\tag 1 \frac {24,158}{6} = 4,000 + \frac{158}{6}$


Now $60$ goes into $158$ but $600$ is no good. Also, $6 \times 2 = 12$.

$\frac {158}{60} = 2 + \frac{38}{60} \text{ iff }$
$\tag 2 \frac {158}{6} = 20 + \frac{38}{6}$


Now $6$ goes into $38$ but $60$ is no good. Also, $6 \times 6 = 36$.

$\tag 3 \frac {38}{6} = 6 + \frac{2}{6} \; \; \text{ Division 'breakdown' Complete! }$

Combining (1) and (2) we get

$\tag 4 \frac {24,158}{6} = 4,000 + 20 + \frac{38}{6} = 4,020 + \frac{38}{6}$

Combining (4) and (3) we get

$\tag 5 \frac {24,158}{6} = 4,020 + 6 + \frac{2}{6} = 4,026 + \frac{2}{6}$

This 'carve-out' method is really long division, but you organize your work in a different way.


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