MATHEMATICS  PROBLEM  131

 

          Find all real valued functions  f  such that

 

 

for every real valued function  g.  (This problem was inspired by a student who incorrectly wrote  “”)

   

Each midshipman submitting a correct solution with a correct explanation to Problem 131 by 1700 Wednesday 30 April 2003 will win a cookie.  Submit solutions to Prof. Wardlaw at mathprob@usna.edu (please no attachments!) or via his mailbox in Chauvenet 301.

         

No one submitted a correct solution to Mathematics Problem 130. The problem is a slightly modified version of the 2002 Putnam Problem A4.  The solution is below and is posted on the board.  

  

MATHEMATICS  PROBLEM  130

 

       The odd-even 3x3 determinant game is played by two persons who alternately enter integers in a 3x3 matrix.  Player 1 enters any integer he desires in the empty 3x3 matrix, then Player 2 any integer he desires in a vacant position, and play continues in this way until the matrix is completed with nine integer entries.  Player 1 wins if the resulting determinant is odd and Player 2 wins if the determinant is even.  If both players use optimal strategies, who will win and how?

 

          Comment.  Math Prob 130 is a slightly modified version of the 2002 Putnam Exam Problem A4.  The solution below is a paraphrased version of the solution to the Putnam problem which appeared in the February 2003 issue of  Mathematics Magazine.

 

          Solution.  The Player 2 will always win if he plays well.  The determinant  a11a22a33 + a12a23a31 + a13a21a32 – a11a23a32 – a12a21a33 – a13a22a31  can be made even if each of the six products are even.  The first part of Player 2’s strategy is to make the two products affected by Player 1’s first play even by choosing even integers for factors in each of these products.  Since there are two possible plays for each of the two products, Player 1 cannot prevent this.  Since each factor appears in one product with a plus sign and one with a minus sign, the remaining two products where Player 2 has yet to play are of opposite sign and thus share a factor.  If this factor is available for Player 2’s third play, Player 2 makes the factor even and wins.  Otherwise, in Player 1’s second and third turn, Player 1 played on the shared factor, and perhaps on one other factor in these two products, as well.  But Player 2 can use the third turn to put an even integer into one of these products (The one with only one factor remaining, if there is such.), and still have the last turn left to put an even integer into the remaining product.  Since all of the products are now even, Player 2 wins.