Problem statement
For all odd prime number p greater than 3, there exists a positive integer n, such that p = 6n +/- 1 :
Check this formula for some given odd primes in a vector by computing n for each p.
Examples
  • p = 17 => n = 3;
  • p = 19 => n = 3;
  • p = [5, 7, 11, 13, 17, 19] => n = [1, 1, 2, 2, 3, 3];
  • p = [5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97];
=> n = [1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 9, 10, 10, 11, 12, 12, 13, 14, 15, 16];
Forbidden functions / expressions
  • regexp
  • assignin
  • str2num
  • echo
See also
Solve

Solution Stats

47 Solutions

35 Solvers

Last Solution submitted on Feb 06, 2026

Last 200 Solutions

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Problem Recent Solvers35

泽伟 Benjamin Blake Harsh Mali

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