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I'm a newbie in math and while doing some exercises and reviewing basic operations I got myself thinking about something that is funny, at least.

If you want to divide, let's say, 13570 by 14 what you should do is: start by taking 1 and see if you can divide it by 14. Since you can't, you take the next number which is 3 making 13, also not divisible by 14. Then, you take the next one making 135 which is divisible by 14.

The next step is the one that I would like to know a better solution. You have literally to guess how many times 14 fits in 135 before you can proceed. In this case it would be 9 with 8 as a reminder which would be taken to the next number so you can proceed.

So, is there a way that we can solve this without this "guessing"?

Thanks

Dulce
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    You estimate. It's not really "guessing", but rather a combination of knowing the "small" multiplication tables; for instance, here, you should know that $4\times 8=32$, and so $14\times 8$ will be around $80$ ($10\times 8$) plus $30$ (which is close to $8\times 4$); and $14\times 9$ will be $90$ plus about $35$ (close to $9\times 4$), so $9$ it is. You aren't guessing, you are making "educated estimations". – Arturo Magidin Aug 02 '11 at 19:36
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    Adding on to what Arturo said, note that 13570~10^4 and 14~10, so you would expect the answer to be approximately about 10^4/10 = 10^3. In this particular case you end up with 9XX. – picakhu Aug 02 '11 at 19:48
  • you just subtract and keep track (at least if youre a computer) until you are left with something between 0 and 14: $0\leq13570-14n<14$. no guessing there – yoyo Aug 02 '11 at 20:05

3 Answers3

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For this particular problem it's a lot easier to use the power of subtraction. $13$ may be less than $14$, but it's not a whole lot less. In fact,

$$13570 = 14000 - 430.$$

It's pretty easy to divide $14000$ by $14$, so now all we have to do is divide $430$ by $14$. Here, if you know that $3 \cdot 14 = 42$, then $430 = 420 + 10$, so

$$13570 = 14000 - 420 - 10$$

hence $\frac{13570}{14} = 1000 - 30 - \frac{10}{14} = 970 - \frac{5}{7} = 969 + \frac{2}{7}.$ This is my preferred technique for dividing small numbers into large numbers; it requires that you are very comfortable with your multiplication tables, but it has the benefit of being very fast once you get used to it, even mentally.

Another way of phrasing what I did is that $135$ is close to $140$, and it's easy to divide $140$ by $14$, so we might as well work with $140$ instead.

P L Patodia
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Qiaochu Yuan
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    I think you mean "dividing large numbers by small numbers"; if you are dividing small numbers by large numbers, division with remainder is very easy! (-: – Arturo Magidin Aug 02 '11 at 19:46
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    This reminds me of a time I put the very simple question $253/23=?$ in the countdown round of an 8th grade math contest. A few people were perplexed as to why I would use such a basic question, but I reasoned that those most capable of seeing the broader patterns in elementary arithmetic would be the ones best able to rapidly divide it by seeing $253=230+23$. – anon Aug 02 '11 at 20:11
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This is just me. I do it this way.

I got tired of all of the guessing stuff and just make a multiplication table. I am fairly quick at arithmetic so I can make the table pretty fast. So I get exercise doing multiplication and I make less mistakes doing long division.

I do it vertically, but here it makes more sense to do it horizontally.

  • I start by setting it up

$$\begin{array}{r|r|r|r|r|r|r|r|r} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 14 & & & & & & & & & \end{array}$$

  • By consecutive doubling, I get the multipliers $2,4,$ and $8$.

$$\begin{array}{c|c|c|c|c|c|c|c|c} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 14 & 28 & & 56 & & & & 112 & \end{array}$$

  • I get times $3$ by adding times $1$ and times $2$, $(14 + 28 = 42)$ or by just multiplying by $3$.

  • I get times $6$ by doubling times $3$.

$$\begin{array}{c|c|c|c|c|c|c|c|c} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 14 & 28 & 42 & 56 & & 84 & & 112 & \end{array}$$

  • I get times $5$ by taking half of times $10$. $(140 \div 2 = 70$.)

$$\begin{array}{c|c|c|c|c|c|c|c|c} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 14 & 28 & 42 & 56 & 70 & 84 & & 112 & \end{array}$$

  • I get times $7$ by adding times $3$ and times $4$ ($42 + 56 = 98$).

$$\begin{array}{c|c|c|c|c|c|c|c|c} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 14 & 28 & 42 & 56 & 70 & 84 & 98 & 112 & \end{array}$$

  • Finally, I get times $9$ by adding times $4$ and times $5$ ($56 + 70 = 126$). I usually check by adding times $1$ and checking that I get times $10$ $(126 + 14 = 140$).

$$\begin{array}{c|c|c|c|c|c|c|c|c} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 14 & 28 & 42 & 56 & 70 & 84 & 98 & 112 & 126 \end{array}$$

This may not be the quickest way, but, with practice, it is pretty fast and pretty accurate. With large divisors, I find that I prefer doing it this way.

Steven Alexis Gregory
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Assume the dividend ($13570$ in your example) has hundreds of digits before or after the decimal point. If the divisor $d$ ($14$ in your example) has only finitely many decimal digits then it is enough to have a table of all multiples $d$, $2d$, $\ldots$, $9d$ handy (in your head or on paper), and at each step you can decide on the correct next digit of the quotient without guessing. It is another matter if your divisor has infinitely many decimal places, as in the case $d=\sqrt{2}$. It is possible to cook up examples where you have to look at more and more extra digits in the dividend before you can decide on the next digit of the quotient.

Christian Blatter
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