Need faster division technique for $4$ digit numbers.

I have to divide $2860$ by $3186$. The question gives only $2$ minutes and that division is only half part of question. Now I can't possibly make that division in or less than $2$ minutes by applying traditional methods, which I can't apply on that division anyways.

So anyone can perform below division using faster technique?

$2860/3186$

This is a multiple choice question, with answers $6/7$, $7/8$, $8/9$, and $9/10$.

• You mean reduce the fraction? Or find the decimal-representation? – BlueRaja - Danny Pflughoeft Aug 4 '10 at 13:38
• What is the rest of the question? Quite likely you don't need the exact value of the division for the rest. – ShreevatsaR Aug 4 '10 at 14:01
• Oops, sorry. The options given are 6/7, 7/8, 8/9 and 9.10. – Electrifyings Aug 4 '10 at 14:07
• a/b = c/d iff ad = bc. So you can multiply instead of dividing. – anon Aug 4 '10 at 14:30
• You can see immediately that 3186 is neither divisibile by 10, nor by 8. And 2860 is not divisible by 6 (it is divisible by 2, but not by 3). So you are left with only one choice: 8/9... which is wrong! Did you copy the correct options? – zar Aug 4 '10 at 16:11

if an approximate answer is enough, you may just throw out digits at the right of the two numbers; so 2860/3186 is more or less 28/31 (it's better to leave two digits, especially because 2860 is much more than 2000 while 3186 is not so much more than 3000)

A second kind of approximation consists in adding or subtracting from both sides two numbers whose ratio is more or less what you expect the answer is. If you start from 2860/3186, which is near 1/1, you may subtract 186 from both sides, ending up with 2674/3000 ~ .891; if you are quick in additions and subtractions you may first add 14 to both sides, obtaining 2874/3200, then subtracting 180 and 200 (whose ratio is .9) obtaining 2694/3000 ~ .896.

• Thanks bud, seems to be perfect solution for me. :) – Electrifyings Aug 4 '10 at 14:58

Another very specific trick, based on the multiple choices: since all the choices are of the form $\displaystyle \frac{n-1}{n} = 1 - \frac{1}{n}$, you're trying to find $n$ for which $1/n$ is closest to 1−(your fraction). So you'd consider (denominator−numerator)/denominator = (3186−2860)/3186 which is (around 320)/3186, clearly closest to 1/10.

If this problem appears on a standardized test you may want to take into account some broader considerations... How many of these digits are even significant (relative to other quantities you will be computing with in "second half" of the problem)? What further operations will you be performing with this quantity - i.e. will you be significantly amplifying your error with nonlinear operations?

At this point you may find that you can simply approximate the quantity as $\frac{2.9}{3.2}$, which can quickly be converted to the decimal 0.906, or so. Seeing that the original number is approximately 0.898, you should be good to go.

• Thanks man, yeah I understood it now. :) – Electrifyings Aug 4 '10 at 14:58

If you are given options like those in your comment (ie. where the numerators/denominators are of comparable size) you can use the fact that:

$$\frac{2860}{3186}=\frac{a}{b} \iff 2860b= 3186a$$

To see which $\frac{a}{b}$ is best, calculate $2860b- 3186a$ for each starting from one of the two midsize fractions and repeating for larger or smaller fractions depending on the sign of the result (if your answer is positive a larger fraction would approximate better, if negative: a smaller one). The choice of a and b giving the smallest answer is the correct one.

• The choice of a and b giving the smallest answer is correct in this case, but not generally. Consider 898/1000 as another option. Its closer to the actual value than the others, but gives a larger difference than all of them. – Jens Aug 4 '10 at 14:47
• Considering the time limit, checking all the options would probably take more time. But hey, that ratio technique you applied is something new for me and would work in lots of other places, so thanks a lot. I really appreciate your answers from bottom of my heart. :) – Electrifyings Aug 4 '10 at 15:00
• @Jens, you are absolutely right! As soon as I had turned off my computer I realised I had forgotten to add a caveat.... – Tom Boardman Aug 4 '10 at 17:32

Continued fractions give the best approximations to a number using smaller terms in the fraction. This may take a bit of practice to calculate at speed, but for the number in question there are only 8 terms:

[0;1,8,1,3,2,2,7],


with convergents:

0, 1, 8/9, 9/10, 35/39, 79/88, 193/215, 1430/1593

One can clearly see that the option 9/10 is the best one from the choices on offer. Note the convergents get progressively more accurate and oscillate between being over-estimates and under-estimates. The final term is, of course, the original fraction itself in its simplest form.

Well 3186-2860 = 326. That is very nearly a tenth of 3186, but 3186/9 = 354. But 326 is closer to 318.6 than to 354, so I'd go for 9/10 instead of 8/9.

This isn't so much a math answer as a test-taking answer: You don't have to compute the fraction, you just have to determine which is the right answer.

2860/3186

This is a multiple choice question, with answers 6/7, 7/8, 8/9 and 9/10.

If you reduce the fraction to get a/b, then b must be a divisor of 3186. This allows you to immediately eliminate some choices. It can't be 9/10 because 10 doesn't divide 3186. You can quickly check that 7 and 8 don't divide 3186, but 9 does, so the only one of the choices that has a shot at being the correct answer is 8/9.

Incidentally, none of those answer is correct; it's not 8/9 either. The furthest you can reduce the fraction is 1430/1593, so you must have an error in your question. Either the fraction is wrong or you're supposed to find the best approximation rather than the actual value.

You could maybe use Euclid's Algorithm to find the GCD of these two numbers, and use this to reduce the fraction. With luck, the result will be easier to manage.

• Not a bad idea, but the gcd is 2, which doesn't help much. – Michael Lugo Aug 4 '10 at 13:56
• I had a look at that algorithm, seems like it would work in many other situations, thanks. :) – Electrifyings Aug 4 '10 at 15:01