If there is one
cheating husband, his wife knows that no one else's husband is cheating, so hers must be because there is
at least one cheating husband. She brings him to the
public square on day one.
If there are two cheating husbands, each wife sees the other cheated-on wife not bring her husband to the public square on day one, and therefore concludes that her own husband is
cheating. (The other wife would only bring her husband to the public square if she knew that no other husbands were cheating. Because she doesn't bring him, at least one other husband must be cheating. Because the only
unknown value for the first wife is her own husband, she concludes that he must be cheating.) Both wives bring their husbands to the
public square on day two.
If there are three cheating husbands, each
wife sees the other cheated-on wives not bring her husband to the public square on day one, or day two. Therefore, she concludes that her own
husband is cheating. (The other wives must know someone else is cheating, and that someone else must be her husband.) All three bring their husbands to the public town on day
three.
Thus, given
k cheating husbands, they will all be brought to town on day
k. Assuming, of course, that the wives are inclined toward reasoning things out rather than beating their husbands with a
frying pan until they 'fess up.