353 views

1 Answer

Best answer
2 votes
2 votes

The above problem could be solved using the concept of cross product.

(As far as i have understood

  1. if n=0, it should be rejected as 0 is a multiple of 5.
  2. if n=3, it should be accepted as 3 is a multiple of itself.
  3. if n=5, it would be rejected.
  4. if n=15, then also it would be rejected.)

selected by

Related questions

0 votes
0 votes
0 answers
1
Aryam asked Jun 8, 2023
202 views
Can anyone help me in this exercise:Show that the language L = {an : n is a multiple of three, but not a multiple of 5} is regular with DFA.i make DFA but i don’t sure ...
0 votes
0 votes
0 answers
4