Factoring and Cancelling in Long DivisionDate: 08/29/2003 at 01:19:38 From: Joan Subject: How do I perform long division? i.e. 1,002,240,000 / 86400 I'd like to learn how to do this. I can get the answer with my calculator but would like to be able to do it on paper. Thank you, Joan Date: 08/29/2003 at 11:01:28 From: Doctor Ian Subject: Re: How do I perform long division? i.e. 1,002,240,000 / 86400 Hi Joan, The first thing to do is note that a fraction is the same thing as a division. That is, __ 4 ) 3 and 3/4 are just two ways of writing the same thing. Similarly, ______________ 86400 ) 1,002,240,000 is the same as 1,002,240,000 ------------- 86400 That's useful, because right away we can get rid of some of the zeros: 1,002,240,000 100 * 10,022,400 ------------- = ---------------- 86400 100 * 86,400 10,022,400 = ---------- 864 And at this point, if you want to go ahead and do the division, it looks like _________ 864 ) 10022400 and the first step would be to decide how many times 864 goes into 1002, since that's the smallest string of digits that is larger than 864: 1 _________ 864 ) 10022400 864 ---- 1382 Next, you bring a digit down, and decide how many times 864 goes into 1382, 11 _________ 864 ) 10022400 864 ---- 1382 864 ---- 518 and so on. If you're just learning this algorithm, you might want to take a look at some of the explanations here: http://mathforum.org/library/drmath/sets/select/ dm_long_division.html But that looks like a pretty painful division! And in fact, we can reduce the fraction further, by finding common factors: 10,022,400 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 * 11,600 ------------ = -------------------------------------- 864 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 = 11,600 When we do this, we see that there's no division left to do. All the prime factors of the denominator are also factors of the numerator. So the first point is that you can often use this kind of simplification to get an easier division problem. Note that you don't have to convert to a fraction, if you don't want to. Given something like ______ 2345 ) 67890 you can simply note that dividing both numbers by 5 won't change the result, so you can go ahead and do that: ______ ______ 2345 ) 67890 = 469 ) 13578 and keep doing this as long as you can identify common factors. Also, depending on how accurate your answer needs to be, you can also use estimation, e.g., you might approximate 86,400 by 85,000 or 90,000, if that would be 'good enough for government work'. Speaking of which, having worked at NASA I recognize 86,400 as the number of seconds in a day, which suggests that you're trying to figure out how many days it takes for something to happen a certain time. I'm guessing that you computed the numerator and denominator separately, and combined everything into a single number in each case, because that's what you've been trained to do. But suppose you'd left all those multiplications undone, giving you something like stuff * stuff * 15 * stuff * 4 * stuff ---------------------------------------- (24 hr/day) * (60 min/hr) * (60 sec/min) Then you could do this: stuff * stuff * 15 * stuff * 4 * stuff ---------------------------------------- (24 hr/day) * (60 min/hr) * (60 sec/min) stuff * stuff * stuff * 15 * 4 = ---------------------------------------- (24 hr/day) * (60 min/hr) * (60 sec/min) stuff * stuff * stuff * 60 = ---------------------------------------- (24 hr/day) * (60 min/hr) * (60 sec/min) stuff * stuff * stuff = ---------------------------------------- (24 hr/day) * (60 min/hr) * (1 sec/min) It turns out that there is often a good reason for leaving operations undone until the last possible moment. In a word, it can be summed up as 'cancellation': The easiest division to do is anything divided by itself, so you want to leave yourself opportunities to do that whenever you can. This is one of the few times in life where procrastination actually pays off, so you might as well learn to take advantage of it. I hope this helps. Write back if you have more questions, about this or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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