(Brady: What's today's video
about? What you got here Matt?) - So this is the handwritten results from a
mathematician who lived in the mid to late 1800s, a gentleman called William Shanks, who I
think in the modern day we would call a recreational mathematician. He had
a real job, ran a boarding school; in his spare time he just did calculation
after calculation. William sent all of his incredible results in to the Royal Society - here
we go - so this is W. Shanks' Reciprocals of Prime Numbers. And then what follows is- and these are
the sections and when he did him so you can see. He's got them in batches of ten to five
thousand at a time, and this is across years, this was not a fast process. And then here's
what he found, and he's described it here. In the left-hand columns of this table are
primes, these are the prime numbers down here, in the right-hand columns is the number of figures
in the period of the reciprocal of each prime. So what he's done is he's taken each prime number,
he's computed 1 over that prime number, and then worked out how many digits before it repeats.
So the first one out of the gates is 60,013 and Shanks is claiming after 5,001 digits
its reciprocal will repeat. So 1 divided by 60,013; it will seem random for 5,001 digits then
you're like, this is familiar, it'll be the same as earlier. Should we check one of them? Because
on the way in today I wrote a little bit of Python software - my ShanksBot - which will do to this
automatically. So give me a prime and I'll put it in here and see what it says.
- (I'm going to pick 61,561.) That has 405 digits. What have we got? 405! Shanks and ShanksBot agree.
So I tell- we'll do one by hand and I'll show you what the ShanksBot is doing and that'll show
us what the worst case scenario can be. Okay, so we'll start with the prime 7 which means
the reciprocal of that prime is 1 over 7. And this means 1 divided by 7 or, in primary
school terms, how many times does 7 go into 1? So let's do it as a long division. So here you've
got your dividend, that's the number going into it, and here you've got the number you're you're
dividing it into, so it's 7 into 1. Now the issue is 7 does not go into 1 because - this is
not very groundbreaking - 7 is bigger than 1. We have a problem. However we can try and work
out what fraction of it is going to go in. And so actually while 7 doesn't go into 1, 0.7
does go into 1. So if I say 0.7 goes into 1 once, that's the same as saying 7 goes into 1 zero-point-once. So what we're saying is 7 goes into 1 zero-point-once but it's got a leftover, 0.3. So what we do now is we're going to put our leftover down here. What we've done is we've said if you've got
1.0 and you put in 0.7 what you're left with is 0.3. And now we're going to repeat that. Implied
are all these extra zeros right? Because there's infinitely many of these, which is very convenient;
we just need to know now - we've already covered this bit - how many times 7 goes into that. And
now we're like, okay, we can't do- 7 doesn't go into 0.3, 0.7 doesn't go to 0.3 but 0.07 does. And so you go how many times does 0.07 go in? Well
4 times 0.07 is 0.28, and that's 4 times it so we put it up
here, and then we subtract that off and we've got 0.020. Now at this point we actually don't need
to write in all these zeros every single time. All we really care about is the 2 and the 0
after it there right? And so we go how many times does 7 go into 20? Well it goes in twice, it's
14. Put the 2 up there, subtract that off we get 6. Now we can add on one of these extra spare 0s, of
which there are infinitely many, 7 goes into 50 that's eight times? Put the 8 up there, subtract
that off, it's 4, now it's 40. 7 goes into uh 35 five times, put the 5 up there - and
all I'm doing is the same logic as before but now it's just a bit more algorithmic, it's a bit
more straightforward. 49, 7, 1. Ah! That's familiar, that's what we had before. And so now actually
we've got exactly the same situation down here, where now with 7 going into 10- then before I
was faffing around with the 0.7s - basically what I'll say is 7 goes into 10 once, that
7 has got a remainder of 3, the 1 would go up there. As you can see we've now got exactly
the same here, 30 would have 28 underneath, we'll put another 4 there. And soon as we hit here that
is exactly the same as what we started with here, 1 and a 0, and so this takes us back there. In theory
it- you know, it's carrying on down here but because we've got exactly the same situation all these
steps will follow in the same order every single time, which is why 7 repeats after 142857, and that's six digits.
- (That's when) (it hits this infinite loop? Will all primes hit an infinite loop like this though?)
- Yes and let's do- let's do a slightly bigger one and we'll be
able to see why it has to happen eventually. So brady here's what we're going to do: we're going
to try and find the reciprocal of a bigger prime, let's keep it to two digits, what would you like?
23 that is a prime we're gonna do a long (Umm 23?)
- 23! That is a prime. We're gonna do a long division and we're gonna divide it into 1. Now as
you may have noticed with 7s, each time I have a result I'm trying to work out what multiple
of the dividend goes into it. So in this case straight out of the gates, 1.0 - we're going to| have to
start putting some zeros on. It doesn't go into 1, doesn't go into 10, does go into 100. You're like, well how many times? I think probably 4. And you're like and four times that
is going to be 92, and then we did that four times so this is actually our answer 0.04. We're just
trying to find a number in the 23 times table that is the biggest but not bigger than
whatever our remainder is at the time. So what I'm actually going to do over here, I'm
going to write out my 23 times table if that's okay. So 1 times 23 is 23, 2 times 23 is we just
double that, 46. 3 times 23 is add those together, 69. 4 times 3 is just double what two times
was so that's 92 - which is what we've already used there. 7 times is add three
and four together; so 9 and 2 is 11, 6 and 9 is 15, add 1 is 16.
- (Wouldn't it just be easier to add 23 to that?) Or you could just add 23 to the previous one. Okay I'm going to be I'm going to be honest Brady you're absolutely right; and what I'm
actually doing is um because I've done this on a much much bigger scale when calculating pi
by hand I've gotten used- I was doing this for like 20 digit numbers right? And so I've just gotten
in this formula- you're right I got so obsessed with my silly way of doing it that I
didn't stop and think, is there a better way? Let's carry 1 - okay there it is cool. And if we
add 23 again we'd be up to 230, that's that's ten times. Okay so this is going to speed up filling it
in a lot. Then if we subtract that off that's 8. Okay so now it doesn't go into 8, it does go
into 80. 69 is the biggest one we're going to have, and that's the 3. Subtract that off we're
left with 11, doesn't go into 11, 110? Yeah 92. And that's 4 times up there.
We subtract this off that's going to be 18, doesn't go into 18, 180? Yes, that's
going to be 161 is the next biggest one, 7 times. As you can see once
you get to the swing of this - 19, 190; 184, 8 times. Subtract that off, 6. So now, what
we're looking for is the first time- 60, yeah 46, twice. The first time we have a point where
this remainder is one we've seen before. So we're at 14 now and we've never seen 14 before,
so I've got to do 14. And it's going to be 138. That gives us a remainder of this time just
2. So have we seen 2 before? We haven't, so the numbers we care about are the top ones, which are
the remainders each time. So so far we've seen 100, we've had 80, we've had 110, we've had 180,
we've had 190, we've had 60, we've had 140, well now we've had 200. Then we said that 200 is 184 is
8 times, put an 8 up there, remainder here is 16. Have we seen 160 before? No we haven't
so now we keep going. Now the question becomes how long can we keep going? And
each time we go we add one more here until we have to hit a repeat. Well in this
case, ignoring the padding zeros, if we just take the non-zero bit that's actually 1, 8, 11, 18, 19, 6,
14, 2. Those- that can't be bigger than 23 because if it was bigger than 23 we would have just picked
a bigger number underneath and taken off one more lot of it. So the remainder, ignoring the zeros,
is always smaller than 23. So it can't be zero, it can't be 23, but it can be any of the numbers
in between all the way up to 22 can appear in here. They may not all appear in there, we may hit a loop
point before we get there, or they may all appear in there somewhere. And if we just redid this as
n then the number of options before it is n minus 1. And so that's your worst case scenario, is it's
going to take n minus 1 steps. Once we've done that we've used every single one, the next one has to
be one we've used before, we'll hit a loop point. So funnily enough, we pull out the Shanks chart, we see
here 60,017 that was as bad as it gets. That was 60,016 digits so it hit every
single one of them before it finally repeated whereas the one before it only hit 5,000,
it hit like - what's that? Like you know, less than 10% of the possible values before it hit a loop point.
- (Surely Shanks wasn't doing this for all-) You know what, I don't think he was because -
partly because this takes a really long time, and you can see me doing it. The first thing you've
got to do is you've got to generate your table, and then you've got to go through and do this,
and as you can see the way I'm doing it takes up a lot of paper.
- (And for a small number!) And for a small number. And what you can do when you get here - so I've done a lot of long division of my time -
what what we did was we you want printed paper with squares on it so everything stays lined up,
and then when you get here you copy it up and you can then you do another stripe, and then you copy
it up and you do another stripe, but you're still going a long way that way and you can't imagine
he went 60,016 digits in that direction. That's one clue he didn't do it this way. The other clue
is the type of mistakes he made. Here, he thought it repeated every 30,525 and he corrected it to 61,050 which is exactly twice what he thought he had. If you look at all these corrections he's always
either doubling or halving it which makes me think he's got a technique that probably involves powers
of two or something, or some steps of halving, and sometimes he's done one too many or one too
few and that's what he's corrected. So that kind of systematic error leads me to believe he wasn't
doing it this way, he had another way of doing it. He died in 1882 so this is like three years before
he died, he's still sending in, he's up to primes up to 110,000. So where did he get to on this one? Oh
there's a note at the end - 'Note: we may hear remark that by means of our table we can readily find the
number of figures in the period of the reciprocals of a composite number not having any higher
factors than 110,000' - so he's basically said these are useful because
you can use the reciprocals of the primes to get all the reciprocals of the composites.
Doesn't say why that's useful. 'We have simply defined the least common multiple of several
numbers of figures in the...'. Yeah okay and he explains how to do that, and then he signs off.
His only practical application for what he's done is you can use it to find another
thing with the practical application. I like this guy! He's got- I really- I like the cut of his jib. (Imagine how good his YouTube channel would have been?)
- Ah he definitely would have had a YouTube channel.
- (Oh yeah. What would he have thought of
the ShankBot?) I think- well I don't know because if his sole goal was to find these numbers he
would love ShankBot. I feel like it was the doing that he enjoyed, and I feel like the fact that
I coded it up and ShankBot just kicks it out without flinching, not the same.
- (One last question because I know you've been doing a lot of research) (into him - was he married?)
- Yeah yeah, and his wife survived him by a good uh pretty much two decades. Yeah so he was like- every morning he's like morning honey, I'm off to work. And by work he would just walk upstairs, start doing calculations,
like oh another another hard day in the office. There are some obvious follow-on questions
that you could have at this point. You could ask yourself how do you take the reciprocals - at
least the length of the repeating part of the reciprocals of the primes - and use it to work out
the repeating part the reciprocals of composites, which is what Shanks put in there as a little
note. You can also ask yourself what's special about the numbers like 7 that use every
single one of these before they get back to where they started. And it it turns out it's
something called- you've got a thing called a primitive root. So prime numbers don't have factors
but they can't have primitive roots which involves doing powers of things and remainders. And primes
that have 10 as a primitive root will have every single one of these pop up; because 10's their
primitive root they go through every single one before they get back to the first one. Which is
why 60,013 does not have 10 as a primitive root so it only used 5,001 and then 60,017 does have 10
as a primitive root and that's why it took the took that long, it used every single one before it
got back to where it started. And there's a thing called a Reptend prime which also tells you
the ones that have this incredible property. So if people want to look into it there's a lot more
fun. For me however all about the long division. (So just so I'm clear, 60,017 are Reptend primes?)
- Yes. (Because they go to the max.)
- Yeah yeah yeah I will allow that as a definition. If you like this video you're really gonna
like Matt Parker's new video for Pi Day 2022, it's on his channel standupmaths. And also
Matt, myself, and Keith Moore from the Royal Society talk more Shanks on the Objectivity
channel. There are links here on the screen and down in the video description.
- Hero. Yeah, hero. Northumbrian hero. And still world record holder for
most digits of pi calculated by hand.
