Each issue Jerry, O4R's puzzle guy, brings you a set of puzzles. Up until now you've had to wait until the following issue of Pro Facto to check your answers. Now you will be able to compare your answers with Jer's by checking the Puzzle Zone on our Web page (where you are now). Answers will be posted about two weeks after the newsletter goes out. However, if you have comments, questions or disagreements, you can still contact Jerry.

     PROBLEM 1

     Warden: "Charlie, tomorrow I execute you."

     Charlie: "Oh sir, I implore you to give me a chance to live."

     Warden: "OK., Charlie, I'm just an old softie. Here's what I'll do for you. I'm going to leave you two bags, one containing 50 white marbles, the other 50 black marbles. You can mix them up any way you want. Tomorrow I'll come in blindfolded, select one bag from which I will remove one marble. If it is white, you live; if it's black you die.

     How should Charlie arrange the marbles to optimize his chances to see the day after tomorrow? What is that optimal chance?

     PROBLEM 2

     C.T.: "Knock, knock, I'm the census taker. How old are you?" The lady answers the question and the census taker asks if any other people live in the house and, if so, their ages.

     Lady: "There are three other people who live here. The product of their ages is 1296 and the sum of their ages is this house number." C.T.: (After doing a little figuring) "You didn't give me enough information. Are you the oldest of the four?"

     Lady: "No."

     C.T.: "Thank you."

     How old are the other three inhabitants? (BIG hint: 1296 = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3)

ANSWERS

     1) Charlie puts a white marble in one bag and the other 99 marbles in the other. He has half (1/2) a chance the warden will pick the bag with only one (white) marble and half a chance he'll pick the other bag (in which forty nine ninety ninths (49/99) of the marbles are white. This arrangement gives Charlie (1/2)[1 + (49/99)] = .747474747...... (or nearly 3/4) a chance to avoid execution.

     2) The product condition requires the ages of the three others to be one of [1296, 1, 1], [648, 2, 1], [432, 3, 1], [324, 4, 1], [324, 2, 2], [216, 6, 1], [216, 3, 2], [162, 8, 1], [162, 4, 2],........... There are a total of twenty-nine different possibilities. Of these, exactly two: [81, 8, 2] and [72, 18, 1] have the same sum which must be the house number (or else the answer would be unambiguous) The question about the respondent's age indicates she is between 72 and 81. Her reply determines the ages of the others to be 81, 8 and 2.