Clearly the amounts of free throws Beatrice makes must be maximized to maximize the total number, as 3/5 is greater than 1/3rd. However, Andrew must take a free throw amount that is a positive multiple of three. Beatrice must take an amount that is a multiple of five. We see that the lowest number of free throws that Andrew makes that satisfies the above conditions is 15(3*5, as 105 is divisible by both 3 and 5). Andrew taking 15 free throws and Beatrice taking 90 free throws leads to a total of 5+54=59 maximum successful free throws.