Numbers composed of a series of 1s: like 111=3x37,1111=11X101. How many of these are prime?

Answer 1

01.) 1
02.) 11 = 11
03.) 111 = 3 x 37
04.) 1111 = 11 x 101
05.) 11111 = 41 x 271
06.) 111111 = 3 x 7 x 11 x 13 x 37
07.) 1111111 = 239 x 4649
08.) 11111111 = 11 x 73 x 101 x 137
09.) 111111111 = 3 x 3 x 37 x 333667
10.) 1111111111 = 11 x 41 x 271 x 9091
11.) 11111111111 = 21649 x 513239
12.) 111111111111 = 3 x 7 x 11 x 13 x 37 x 101 x 9901
13.) 1111111111111 = 53 x 79 x 265371653
14.) 11111111111111 = 11 x 239 x 4649 x 909091,
15.) 111111111111111 = 3 x 31 x 37 x 41 x 271 x 2906161

Hm... Well 11 is a prime.
About the rest... every third one (111, 111111, ...) is divisible by 3 so it's not prime...

None of the numbers between 111 and the one with 18 digits are prime.

The next prime is the one with 19 digits, then the one with 23 digits. Then nothing up to 70.

Might get back to this space later and try to prove it.

What do you need this for? (might give me a clue)

Are you trying to get us to prove some open conjecture that mathematicians have been trying for decades to prove it ??? ;-)

Naughty, naughty!

Answer 2

These numbers are called repunits. Here are the
known primes in this sequence: those with
2, 19, 23, 317, 1031, 49081, 86453, and 109297 digits.
Incidentally,
Knashha mentioned that it is known that
it is known that no repunit is a square,cube or
5th power.
Well, it's clear that none can be a square because
all repunits end in 11, so are congruent to 3(mod 4).
As for higher powers,
A very recent paper(really a note) on this topic is by Bugeaud and Mignotte:
Sur l'equation diophantienne (x^n-1)/(x-1)= y^q, II (1)
Comptes Rendus Acad Sci. Paris Mathematiques 328(1999), 741-744.
Among other things, they outline a proof of the result that eqn (1) has no
solutions for x=10. In particular, no repunit can be a power!

Answer 3

Well, it seems we can relate this to wilson' prime number theorem, which states that a number p is prime if and only if (p-1)! is congruent to -1 (mod p). Obviously these 111... numbers are all going to be a power of 10 when you subtract 1. Therefore their factoral will also be a power of 10. in order for them to be congruent to -1(mod p) it would be necessary for the largest divisor of (p-1)! by p to also be divisible by 9, as this is the only case where the 1 in the one's digit to be 9, which is the only case where you will have a congruency to -1(mod p). So this offhand will drastically decrease the amount of time you could test for the primality of such numbers.
I feel like if these numbers all besides 11 are in fact composite then a proof culd somehow be derived from this wilsons theorem combined with the fact that all p-1 = (2*5)^n, but I dont know how right now.

Answer 4

Only 11.
111=3x37
1111=11X101
11111=41x271
111111=11x10101
1111111=(couldn't find this one)
11111111=11x1010101

every other is divisible by 11 and 101(repeating minus one from the original number(1111111). Prime numbers are tricky to work with. If you noticed, all the odd ones factors, 111, 11111, are all prime, so this leads me to believe that there is a pattern in there somewhere.

Answer 5

These numbers are called repunits and can be notated and written as

Rn = (10^n-1)/9.

Only a handful of prime repunits are known:R2,R19,R23,R317,and R1031 plus possibly a few more.

The size of the repunit set is unknown. It has been shown that no repunit is a square,cube, or fifth power.

Answer 6

Well just to give a few observations:
1) if there is a multiple of 3 digits, it's not prime because it will hten be divisible by 3.
2) if there is a multiple of 2 digits (higher than 11), it's not prime because it will hten be divisible by 11.