Talk:Deceptive numbers

Since the repunit ends in 1 5 can not be a factor of n--Nigel Galloway (talk) 11:44, 12 February 2022 (UTC)

Since the repunit is odd, 2 cannot be a factor of n. And because the digit sum of the repunit is n - 1, 3 cannot be a factor of n (an integer is divisible by 3, if its digit sum is divisible by 3, but n - 1 and n cannot both be divisible by 3).
All non-primes below 7² are divisible by 2, 3, or 5. So it is sufficient to start the search at 49. --Querfeld (talk) 14:01, 24 September 2023 (UTC)

Regarding the repunit R_n-1, the equation R_n-1 * 9 = 10^n-1 - 1 is fulfilled. And if it is ensured, that n is not divisible by 3, then if R_n-1 * 9 is divisible by n, also R_n-1 is divisible by n. In that case, the check of R_n-1 mod n = 0 can be simplified to 10^n-1 mod n = 1, which can be calculated via modular exponentiation, so that big integers are not required for this task.

Yep, that certainly works! (actually nine entries were already doing just that last year) --Petelomax (talk) 23:52, 26 September 2023 (UTC)

With the wp:Fermat primality test being a^n-1 mod n = 1, where gcd(a, n) = 1, it becomes obvious, that the deceptive numbers are a subset of the Fermat pseudoprimes to base a = 10, with the multiples of 3 filtered out (10^x - 1 already rules out the multiples of 2 and 5). --Querfeld (talk) 19:09, 26 September 2023 (UTC)