Transcript Carom 1-8: Repunits - s253053503.websitehome.co.uk

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Activity 1-8: Repunits
111, 11111, 11111111111, 11111111111111
are all repunits.
They have received a lot of attention
down the years.
In particular, when are they prime?
R1 = 1, no, R2 = 11, yes, R3 = 111, no...
Task: prove that any repunit
having a composite (non-prime) number of digits
must be composite.
111111111111111
= 111111111111111
= 111  1001001001001
A common move with repunits
is to consider their value in bases other than 10.
The above argument is exactly the same
in bases other than 10.
So for a repunit to be a prime in base 10 is rare.
Conjecture: there are infinitely many
repunit primes in base 10.
When is a repunit square?
Well, 1 is a square –
but if we search for others, they seem hard to find.
Conjecture: 1 is the only square repunit.
Task: how could we prove this?
How to approach this?
Firstly, what remainders can a square have if you divide by 4?
(2n)2 = 4n2, and so has remainder 0,
while (2n+1)2 = 4n2 + 4n + 1, and so has remainder 1.
Conclusion: a square can never
have remainder 2 or 3 when divided by 4.
Now consider 111...111.
1 is a square.
11 is not a square.
For repunits bigger than these,
We have 111...111 = 111...11100 + 11.
4 goes into 111...11100,
and leaves a remainder 3 when it divides 11.
So 111...111 cannot be a square,
and 1 is the only square repunit.
And so we have...
Theorem: 1 is the only square repunit.
Another theorem: if a number is not divisible by either 2 or 5,
then some multiple of this number must be a repunit.
3  37 = 111
11  1 = 11
7  15873 = 111111
13  8547 = 111111
17  65359477124183 = 1111111111111111
19  5847953216374269 = 111111111111111111
215291=111111
How to prove this?
We can use a theorem due to Euler.
Leonhard Euler,
Swiss
(1707-1783)
The numbers a and b are COPRIME if gcd(a, b) = 1,
where gcd = ‘greatest common divisor’.
Define (n) to be the number of numbers
in {1, 2, 3…, n - 1} that are coprime with n.
So (2) = 1, (3) = 2, (4) = 2, (5) = 4,
(20) = 8.
Task: find (5), (9), (45).
What do you notice?
It turns out that (n) (the totient function)
is what we call multiplicative.
That is to say, if a and b are coprime, then
(ab) = (a) (b).
Thus (45) = (9) (5).
Now Euler’s Theorem tells us:
if a and b are coprime, then a divides b(a) – 1.
So, for example, since 11 and 13 are coprime,
11 divides 13(11)  1,
137858491848 = 11 x 12532590168
and
13 divides 11(13) 1,
3138428376720 = 13 x 241417567440
Note that a repunit is of the form (10n – 1)/9.
Now suppose a and 10 have no common factor.
Then 9a and 10 have no common factor.
So by Euler’s Theorem, 9a divides 10(9a)  1.
So 10(9a)  1 = 9a  k, for some k.
So a  k = (10(9a)  1)/9, which is a repunit.
Thus some multiple of a is a repunit.
Are there any numbers that are repunits
in more than one base?
31 = 111 (base 5) = 11111 (base 2)
8191 = 111 (base 90) = 1111111111111 (base 2)
Goormaghtigh Conjecture: these are the only two.
With thanks to:
Shaun Stevens, for his help and advice.
Wikipedia, for another excellent article.
Carom is written by Jonny Griffiths, [email protected]