# Puzzle Corner

Since this is the first issue of a new academic year, let me once again review the ground rules. In each issue I present three regular problems (the first of which is normally bridge-related) and one “speed” problem. Readers are invited to submit solutions to the regular problems, and two columns later (not every issue of *TR* contains a “Puzzle Corner” column), one submitted solution is printed for each regular problem. I also list other readers who responded. For example, solutions to the problems you see below will appear in the March 2004 issue, and the current issue contains solutions to the problems posed in the May 2003 issue.

I am writing this column in mid-July, and I anticipate that the March 2004 column will be due in December. Please try to send your solutions early to ensure that they arrive before my submission deadline. Late solutions, as well as comments on published solutions, are acknowledged in subsequent issues in the “Other Respondents” section. Major corrections or additions to published solutions are sometimes printed in the “Better Late Than Never” section, as are solutions to previously unsolved problems.

For speed problems, the procedure is quite different. Often whimsical, these problems should not be taken too seriously. If the proposer submits a solution with the problem, that solution appears at the end of the same column in which the problem is published. For example, the solution to this issue’s speed problem is given below. Only rarely are comments on speed problems published.

There is also an annual problem published in the first issue of each year, and sometimes I go back into history to republish problems that remained unsolved after their first appearances.

**PROBLEMS**

**Oct 1.** Larry Kells wants to know, What is your best chance to make 7 spades with

The opening lead is a spade, RHO following. Assume there are no inferences to be had from the bidding or the lead, and that the opponents will make no mistakes for the rest of the play.

**Oct 2.** Donald Aucamp offers us the “Three Hat Problem.”

Three logical people-A, B, and C-are wearing hats with positive integers painted on them. Each person sees the other two numbers, but not his own. Each person knows that the numbers are positive integers and that one of them is the sum of the other two. A, B, then C take turns in a contest to see who can be the first to determine his number. In the first round, A, B, and C all pass, but in the second round, A correctly asserts that his number is 50. What are the other two numbers, and how did A determine his was 50?

**Oct 3.** Fred Gardiol wonders how many different resistances he can obtain by connecting 10 one-ohm resistors.

**SPEED DEPARTMENT**

Some more standard conversion factors from Sanjay Palnitkar. As an example, 1,000,000 aches is one megahertz. What are

One million billion piccolos

10 rations

100 rations

.5 large intestines

The time between slipping on a peel and smacking the pavement

365.25 days of drinking low-calorie beer

**SOLUTIONS**

**May 1.** The following problem is from Larry Kells, who writes that his friend’s latest bridge story about his wife convinced Larry that a lucky charm must hover over their marriage: even the bad turns out to be good.

In one recent game, when vulnerable, he picked up a hand that left him dreading that his luck had run out. He had the worst hand imaginable:

His left-hand opponent dealt and opened 2 hearts, weak. Partner doubled for takeout, and inwardly he shuddered with panic, wondering how he could avert disaster. To his relief, his right-hand opponent jumped to 4 hearts, taking him off the hook. Around to partner: to his shock, she leaped all the way to 7 spades! Naturally this was doubled. Down came the ace of clubs on his right, and my friend sheepishly spread his pathetic dummy, certain of impending doom.

Then, suddenly, everything entered the twilight zone. His true love gushed, “Wow! What a magnificent dummy! I was speculating with the 7 bid. With some hands you might have held, it would have been nearly hopeless, but now our chances for making it are greater than 97 percent.” And indeed, the grand slam rolled home, despite the fact that she held only 16 points, and none of the opponents’ suits broke evenly. Any idea what the full deal might have been?

Robert Bishop found that the friend’s wife could hold

with 16 high-card points (which the proposer must have meant).

The beauty of the friend’s seemingly abysmal dummy is that it contains as many as three trumps and exactly three diamonds.

These holdings imply that (1) the opponents have just two trumps and five diamonds between them; (2) since no suit breaks evenly, one opponent must hold both trumps, so two rounds are needed to draw them; (3) dummy can now ruff a fourth diamond, hopefully setting up declarer’s fifth diamond as her 13th trick; and (4) the grand slam is made unless one opponent improbably holds all five diamonds.

The chance of success when the dummy first appears turns out to be 98.3 percent-even better than claimed in the problem.

Given the six-four split of the opponents’ hearts (from the bidding), there are three possible paths to victory: (1) each opponent holds one trump (since the actual trump division is not yet known), (2) the opponent with six hearts holds both trumps, and (3) the other opponent holds both trumps. The probabilities of (1), (2), and (3) are, respectively, .525, .175, and .3.

With (1), victory is assured because dummy can ruff both of declarer’s two low diamonds. With (2) and (3), the probabilities of victory are, respectively, .9366 and .9790. Furthermore, .175 (.9366) = .164, and .3 (.9790) = .294; so .525 + .164 + .294 = .983.

**May 2.** Here is a magic-square problem from Jon Sass. It is quite different from any I have seen before. You are to construct a four-by-four array of positive integers so that 24 different four-integer sums total 264. This is not so amazing, especially because the four numbers need not be in a row, column, or on the diagonal. If you turn the magic square upside down, again there are 24 four-number groupings that total 264. Note: turning upside down really means turning the paper 180 degrees in the plane of the paper.

Richard Hess exceeded the requirement and found a square with 29 different sums that add to 264. The rotated square has the same entries, though not necessarily in the same locations:

Charles Morton shows us how he developed his solution. A few of the stages are depicted below, and readers who can process .xls files (normally produced by Microsoft Excel) can view and play with Morton’s complete solution on my “overflow” Web site: allan.ultra.nyu.edu/~gottlieb/tr. I don’t run Excel, but the file does open correctly with OpenOffice under Linux.

Aesthetically pleasing corners and the rest of the “1x” members:

Complete three quartets:

Add 69 and 96 to the second diagonal since it’s the only remaining pair that sums to 165:

Complete the square:

Mapping the digits 1, 6, 8, 9 to 0, 1, 2, 3 yields another magic square, which, if subjected to a base-4 to base-10 conversion, makes yet another magic square with the numbers 015.

**May 3.** A celestial mechanics problem from Kern Kenyon: Consider Puzzle Corner as two large-equal-mass points that revolve about their center of mass. Place a third very much lighter mass point collinear with the first two, but outside (not between) them in such a way that the three masses remain on the line with constant separations as the line rotates about the center of mass. Find the distance of the small mass from the center of mass.

The following solution is from John E. Prussing: For the two equal large masses to orbit at a constant distance R from their center of mass C, a balance of centrifugal acceleration and gravitational acceleration gives

the orbital angular velocity. The small collinear mass must orbit at a constant distance x from point C at this same angular velocity. Balancing centrifugal and gravitational accelerations on it gives

Defining y as x/R and doing some algebra gives

y^{5} 2y^{3} 8y^{2} + y 8 = 0. This equation has one real solution, equal to 2.3968.

**BETTER LATE THAN NEVER**

There were no major corrections or additions this month.

**OTHER RESPONDENTS**

Responses have also been received from D. Bator, G. Blum, C. K. Brown, D. Church, H. I. Cohen, D. Dechman, R. Giovanniello, H. Hindman, S. Kanter, K. L. Rosato, H. Sard, R. Schweiker, E. Signorelli, B. Simon, W. Sun, T. Terwilliger, and A. E. Zeger.

**PROPOSER’S SOLUTION TO SPEED PROBLEM**

One gigolo, one decoration, one C-ration, one semicolon, one bananasecond, one lite year.

Send problems, solutions, and comments to Allan Gottlieb, New York University, 715 Broadway, 7th Floor, New York NY 10003, or to gottlieb@nyu.edu.