Long ago in
the dim reaches of grade five or so (in my case it was still called standard
three) we learned short division. For
those of you who don’t remember grade five (or haven’t gotten there yet). I’ll
outline how this works below. Suppose I want to divide 629 by 4. (Obviously
skip this if know it).

Start out by
writing

I can’t
divide this all in my head but I do know that 4 goes into the first digit (6 in
this case) once and leaves a remainder
of 2. So I write.

The next
step is noticing that 4 goes into 22,
five times and leaves a remainder of 2. So I write

Repeating
this procedure a few more times (and adding in some zeros after the decimal
point) gives us.

Which is to
say that 629 divided by 4 is 157.25.

The
refresher out of the way let’s try to divide 1 by 7. Start here.

Now fill in
some of the rest of it (actually do it if you haven’t).

You’ll get
this after 6 iterations.

From here we
notice, that we’re pretty much where we started.

We have to
divide 10 by 7 in both of the previous 2 pictures and we have an infinite string
of 0s in front of us. So we’re going to get the same thing over and over and
over again!!!

So 1/7=0.14285714285714285714285714…….

Just
remember those six digits in order (less than a phone number and you remember
yours) and you can pull off the trick of dividing one by seven.

It gets
better though. Dividing two by seven is just as easy, in fact we have already
done it!!

How you ask.
Well start here

We’ve been
here before? Dividing 20 by seven. We get the SAME repeating decimal with a
different start point (not to beat anyone over the head with anything but
yeah,3/7, 4/7,5/7 and 6/7 work the same way).

i.e.

2/7=0.2857142857…

3/7=0.42857142857….

4/7=0.57142857….

5/7=0.71428571428571....

6/7=0.85714285714…..

All you need
to do is remember your start point (and the 6 digit string 142857) and you can divide by 7.

If you want
to divide say 27 by 7 then. 27/7=3 remainder 6 so 3.857142857…..

This
a pretty neat trick but even better there really
isn’t anything special about 7. Provided we're willing to learna 16 digit string we can do division by 17. Division by 13 requires learing two 6 digit strings

This works for every positive integer which doesn’t have 2 or 5 as a factor (why 2 and 5 because they are 10s prime factors and we're using base 10, yes this also works in every other base for exactly the same reasons).

For example consider dividing 1 by 349. There are only 348 possible things to carry over when dividing 1 by 349 so when I go through the procedure of short division I’m guaranteed to get back to 1 (the first thing I carry) in at most 348 steps and things will repeat. The period might not be 348 but it won’t be more. For example the next 2 “interest numbers” to divide by in base 10 are 13 and 17.

This works for every positive integer which doesn’t have 2 or 5 as a factor (why 2 and 5 because they are 10s prime factors and we're using base 10, yes this also works in every other base for exactly the same reasons).

For example consider dividing 1 by 349. There are only 348 possible things to carry over when dividing 1 by 349 so when I go through the procedure of short division I’m guaranteed to get back to 1 (the first thing I carry) in at most 348 steps and things will repeat. The period might not be 348 but it won’t be more. For example the next 2 “interest numbers” to divide by in base 10 are 13 and 17.

1/17=0.05882352941176470588235294….

This is of
length 16 and provided you can remember where to start you can figure out k/17
for pretty much any integer k. On the other hand

1/13=0.0769230769230769230…
has period 6. So it only gets half of the the k/13s you could wish for.

2/13=0.153846153846….
also has period 6 and gets us the other half.

So when do
we get everything from just 1 cycle??? The answer comes oddly enough from
geometric series.

To see this we'll consider some examples.

1/7=0.142857142857142857….=

142857[(10^-6)+(10^-6)^2+(10^-6)^3+(10^-6)^4+….]=

142857*(1/999999)

Giving us
7*142857=999999.

Same thing
goes through with 13. What this means is that

i.e. 13*076923=999999

The only way to have a period of length (a factor of) 6 is to
be divisible by 10^6-1.

Clearly 17 does divide 10^16 - 1 but none of 10^8 - 1, 10^4 - 1, 10^2 - 1 and 10^1 - 1. So it’s period is actually 16.

There are other ways to think about this and it’s quite abit
of fun to do so. But this post is getting long so I’ll stop here.

Maybe a silly question, but what's an efficient way to determine the period of 1/n in base 10?

ReplyDeleteI realize that the period of 1/n is k implies that 10^k = 1 (mod n), but I'm not sure how to cleanly find the smallest k for which this is true.

"If you want to divide say 27 by 7 then. 27/7=3 remainder 6 so 3.857142857….."

ReplyDeleteOkay, sure, but first you need to get that 27/7 is 3 remainder 6, which is the real meat of the division. This just lets you turn the "remainder" bit into a string of digits, which is neat I guess but doesn't do you a lot of good if you're trying to divide a four-digit number by 7. I dunno, from the post title I guess I was expecting a neat divisibility trick for 7 or something.

Ok so if you want to test for divisibility by 7 then you can cut off the units coloum and subtract it from the other bit twice.

Deleteexample to see if 123456 is divisible y 7

123456 goes to

12345-2*6=12333 goes to

1233-2*3=1227 goes to

122-2*7= 108 which isn't divisible by 7 so 123456 isn't either

Ooh, clever. You're basically subtracting 21 times the units digit from the original number, then dividing the result by 10. Since 7|21 and 10 is coprime to 21, that works. Presumably you could extend this to other divisors, e.g. subtract the units digit 5 times instead of 2 times to test divisibility by 17? (Though 17's pretty easy anyhow because it divides 102 and 1003.)

DeleteI like that way of thinking about it. I guess we need a multiple of 17 which ends in 1 so, cut off last digit and subtract 5 times that from the non-units part?

DeleteM(¹³)+A(¹)+T(²ยบ)+H(8)=42db(Date of Birth)

ReplyDeleteThe result of the math literary equals the last number 4•2=8HeArt or h8.

Also the numbers 42 mirror is 24(x) for the hours of day(times).

This number is also found in the Magikal Sun Table.