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

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.

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.

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).