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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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