SOLUTION: In preparation for Halloween, three married couples, the Browns, the Joneses and the Smiths, bought little presents for the neighborhood children. Each bought as many identical pre

1.  From the condition, it is clear that the amounts each person spent for presents, each (amount) is a perfect square of cents.



2.  If x and y  are spendings for some (any) of the three couples, then

    x^2 - y^2 = 75,    according to the condition,  or

    (x+y)(x-y) = 75.   (and we remember that the greater value is the wife's spending !)


    For integer positive  x and y  it gives these and only these opportunities


        x + y = 75
        x - y =  1     with the solution  x= 38,  y= 37  
    OR
        x + y = 25
        x - y =  3     with the solution  x= 14,  y= 11
    OR  
        x + y = 15
        x - y =  5     with the solution  x= 10,  y= 5.


    Of these solutions,  the only pair  (38,37)  has the difference of 1,  so it gives a clue to me to conclude that

                         Ann is the wife to Bill Brown.



3.  But in order for to make this conclusion ABSOLUTELY CORRECT, I must EXCLUDE that the other couples fall in the same pair/solution (38,37).

    Fortunately, the condition gives me rationality to make this conclusion.

    Indeed, it says that "Betty bought one less present than Joe Jones", which means that Betty is of the pair (10,5), while Joe Jones is 

    of the pair (14,11).  So, I really can conclude that Ann is the wife to Bill Brown.



4.  Finally, from the condition, it is also clear that Betty IS NOT  the wife to Joe Jones.



5.  It leaves only one opportunity  for Mary to be Mary Jones.