31, 331, 3331...

31 is a prime, so is 331 and 3331.
Prove (or disprove) that every number with threes and ending with a one is prime.

Here are some composite numbers of this form:

(10 ^ 9 - 7) / 3 = 17 x 19607843
(10 ^ 10 - 7) / 3 = 673 x 4952947
(10 ^ 11 - 7) / 3 = 307 x 108577633
(10 ^ 12 - 7) / 3 = 19 x 83 x 211371803
(10 ^ 13 - 7) / 3 = 523 x 3049 x 2090353
(10 ^ 14 - 7) / 3 = 607 x 1511 x 1997 x 18199
(10 ^ 15 - 7) / 3 = 181 x 1841620626151
(10 ^ 16 - 7) / 3 = 199 x 16750418760469

Originally posted by David113
333333331 = 17 x 19607843

Correct.

31, 331, 3331, 33331, 333331, 3333331, 33333331 are primes, 333333331 is not.

Once, before computers and mechanical calculators, there was a conjecture that every number started with a number of threes, ending with a one was a prime until, with great effort, 333333331 was factorized, and therefore, the conjecture was disproven.

Originally posted by FabianFnas
Correct.

31, 331, 3331, 33331, 333331, 3333331, 33333331 are primes, 333333331 is not.

Once, before computers and mechanical calculators, there was a conjecture that every number started with a number of threes, ending with a one was a prime until, with great effort, 333333331 was factorized, and therefore, the conjecture was disproven.

Interesting.

I find this hard to believe, since it is easy to prove - without any computer - that this sequence contains a composite number.

By Fermat's Little Theorem, if p is a prime and x is an integer not divisible by p, then x ^ (p - 1) - 1 is divisible by p.

Let p = 31, x = 10. Then you get the result - 999999999999999999999999999999 (30 9's) is divisible by 31.

This means that also this number divided by 3, which is 333333333333333333333333333333, is divisible by 31; and so, 33333333333333333333333333333300 also is divisible by 31; and
so is 33333333333333333333333300 + 31 = 33333333333333333333333333333331.

so 33333333333333333333333333333331 is not prime.

Ah, the old divisibility rules are still going strong

Originally posted by David113
I find this hard to believe, since it is easy to prove - without any computer - that this sequence contains a composite number.

By Fermat's Little Theorem, if p is a prime and x is an integer not divisible by p, then x ^ (p - 1) - 1 is divisible by p.

Let p = 31, x = 10. Then you get the result - 999999999999999999999999999999 (30 9's) is divisible by ...[text shortened]... + 31 = 33333333333333333333333333333331.

so 33333333333333333333333333333331 is not prime.

And I thought im geek...

I came across this puzzle last week;

The sum of the digits of the number 37 is 10, wich has sum of digits 1.

The number 37^2=1369 has sum of digits 19, wich has sum 10, wich results to 1.

A) Prove or disprove that every power of 37 will end in 1 after taking the sum of the digits repeatedly.

B) Prove or disprove that repeatedly taking the sum of the digits of 37^n will be 10, before becoming 1, for all integers n>0.

B implies A, I know, but A is a bit easier then B.

A follows from the fact that 37 = 1 (mod 9), so 37 ^ n = 1 (mod 9).