The answers to the 36 "quickie" type problems brought more surprises by mail than any previous collection of short problems. Readers caught ambiguous phrasings, indulged in amusing quibbles, found alternate and sometimes better answers, spotted some errors, and argued that the last problem is meaningless. I shall comment on this correspondence, taking the problems in numerical order, and add some further observations of my own.

(4) C. C. Cousins, Charles W. Bostick, and others noticed that four of the court cards in the illustration for the answer to this problem are incorrectly drawn. The Jack of Diamonds and the Jack of Clubs should be one-eyed, and the King of Spades and King of Hearts should face the other way. Some readers thought the Queen of Diamonds should face the other way. But Bostick took the trouble to examine 30 different decks made in the United States and found that in 18 of them the Queen of Diamonds faced right, and in 12 cases she faced left, so this card cannot be considered wrong.

(5) This theorem is related to a paradox of induction that I came across in Karl Popper's Conjectures and Refutations where he attributes it to J. Agassi. "All events occur before the year 3000." Since this statement has so far been confirmed by every event in the history of the universe, some theories of induction are forced to regard it as strongly confirmed, thus suggesting that it is highly probable the world will end before 3000.

(8) Larry S. Liebovitch, instead of using a decimal point, solved this problem by using "In," the symbol of "natural log of." Thus 21n3 = 2.19 + .

(11) Problem E 2262, in The American Mathematical Monthly (November 1971, pages 1021-1023), By G.J. Simmons and D. E. Ra~wlinson, generalized this question by asking for all other sets of k positive integers of which the same statement could be made. It turns out it can be made for all positive integers, but only a very small set have unique answers. When k = 2, the only answer is 2 + 2 = 2x2. Our problem provided the only answer for A = 3. For ä = 4 it is 2 + 4+1 + 1= 2x4x1x1.

Readers of the periodical showed that for all values of h not exceeding 1,000, the only values with unique solutions are 2, 3, 4, 6, 24, 114, 174, and 444. It is possible, the editor comments, that no other values have unique answers other than the eight listed.

(12) James A. Ulrich was the first to argue that the probability of the string's being knotted is 1 because there is no way a closed loop of string can exist without its ends being tied.

(13) "House" remains the best answer, but less familiar words such as "ye" and "el" allow other solutions. George A. Miller sent a computer printout of 269 alphabetized answers, and all the words (from "abhor" to "wavey") are found in standard dictionaries.

(14) Martin Kruskal provided a photocopy of the New York Times account (February 22, 1938) of Samuel Isaac Krieger's preposterous claim to have disproved Fermat's last theorem. He had saved the clipping since he had seen it as a small boy.

(15) Solomon W. Golomb proposed "underfund" and "un-derwound" as alternate answers.

(16) The probability of 1/2 that a distant viewer will see three sides of the Pentagon is correct only as a limit as the viewer's distance from the Pentagon building approaches infinity. My solution ignored the fact that there are five infinitely long strips, each crossing the building, inside of which both the viewer and his Doppelgänger can see only two sides (and if very close to the Pentagon, only one side). This was pointed out by readers too numerous to list.

The probability is zero, commented P. H. Lyons, "if the smog in Washington is anything like what it is here in Toronto."

(17) Walter C. Eberlin and David Dunlap independently added two strokes to 11030 so that when it is viewed in a mirror it spells "peon," a word closely related to "hobo" in meaning.

Richard Ellingson took advantage of the fact that I did not specify that the lines of 11030 could not be rearranged. His solution was: ^

(20) Hans Marbet, of Switzerland, pointed out that if ABCD are replaced by any four consecutive digits, and the four dots replaced by the same digits in the order CDAB, there is a valid solution for a number system with the base A+B+D. For example, in base 16:

### 4567 7654 6745 12300

(21) A. P. Evans, William B. Friedman, and others wrote to say that you don't need to know that all primes greater than 3 have the form of 6n plus or minus 1. Only four parallel diagonals can be drawn through the snake. Call them, top to bottom, A,B,C,D. All numbers on A are divisible by 3 and therefore cannot be prime. All on B and D are divisible by 2, and hence cannot be prime. Therefore all the primes must fall on C. "I don't suppose it would be sporting," Evans adds, "to ask readers to come up with a diagram on which a straight line can be drawn that contacts all prime numbers and only prime numbers."

(23) When George said that Feemster "owns fewer than that," I meant him to mean fewer than the amount specified by Albert. If "that" is taken to refer to "a thousand" instead of "more than a thousand," however (as many readers pointed out), Feemster could own exactly 1,000 books as well as none.

(24) Howard J. Frohlich passed along a friend's view that a date such as 8/8/71 could also be called "ambiguous" because you do not know whether the first 8 refers to the day or the month.

(25) Since I failed to ask why manhole covers and holes are round instead of square, scores of readers sensibly replied that the covers are round to fit the holes. John W. Stack cited as his authority for this answer M. A. Nhole's Comprehensive Review of Equilateral Rectangular Beams and Circular Receptacles, pages 31— 4207, published in 1872 by the Sewer and Street Company, Inc. P. H. Lyons had another answer: To reduce the decisions a sewer worker has to make in replacing the cover.

Some covers and holes are square, according to John Bush, who told of a recent explosion near his home in Brooklyn that blew off a Consolidated Edison square manhole cover. After the smoke cleared the cover was found at the bottom of the manhole. "Geometria invincibilis est," Bush concluded.

(28) The Hamlet rebus, "To be or not to be," was invented by Golomb, a fact I did not know when I gave it.

Jim Levy wrote to say that strictly speaking the symbol for "or" should be one that represented exclusive disjunction (either but not both) rather than inclusive disjunction (either or both), otherwise the statement implies that a person can be and not be simultaneously.

(32) "Can you answer this?" Golomb wrote in 1971. "No, U Thant!"

(33) The problem of the touching cubes was one I thought of several years ago and had answered with 20 cubes. I was staggered to receive two different solutions, each with 22 cubes. Figure 46 shows how five white cubes can abut the top side of the grey cube. Since none extends beyond line AB, this formation can go on four sides of the grey cube [see Figure 47].

Figure 46

Figure 46

Two more cubes plug up the holes on face A and its opposite side. This solution was first received from Kenneth J. Fawcett, Jr., and later from Rudolf K. M. Bergan, Michael J. and Alice E. Fischer, Leigh Janes, K. B. Mallory, Allen J. Schwenk and George Starbuck.

Arrangement for 22-cube solution

The other solution, found independently by Bergan, Rudolph A. Krutar and Robert S. Holmes, is shown as drawn by Holmes [Figure 48], Eight cubes go on two opposite faces of the grey cube, and six abut the grey cube in the middle layer. Even the fact that as many as eight nonintersecting unit squares can overlap one unit square is, as far as I know, a previously unknown result.

Stanley Ogilvy later pointed out that because the bottom corners of the three lowest squares in Figure 46 are not on a horizontal line, there is just enough room below them to permit three more squares, joined face to face, to go beneath the other five squares. This provides another way for eight cubes to abut one face, and leads to another solution with 22 cubes.

While I was still recovering from the 22-cube solution, Holmes (who is working for his doctorate in particle physics at the University of Rochester) delivered the knockout punch: a 24-cube solution! Later Janes, in collaboration with Michael Bradley, reported a 23-cube solution.

It is hard to believe, but as far as I know no one has seriously considered before the simple question of how many unit cubes

can share a positive surface area with a given unit cube. My innocent answer of 20 is indeed the best if one adds the condition that the surface of the given unit cube (we shall distinguish it from the others by making it grey and leaving the other cubes white) be completely covered by the touching cubes. This proviso, however, was not part of the original problem.

Holmes's technique begins with placing seven white cubes on one grey face [see Figure 49], Three pairs of cubes (think of each pair as being glued together) are placed around a face of the grey cube so that the midpoint of each pair touches a corner of the grey face. A seventh white cube (P, drawn with broken lines) overlaps the grey face as indicated, the two faces having an axis of symmetry shown as a diagonal line. By rotating the white cubes clockwise, keeping corner A on the left edge of the fixed grey face and preserving the bilateral symmetry, the pattern shown in Figure 50 is reached. If the broken-line cube P is now moved up a trifle, the two meeting corners of each pair of glued cubes can have a tiny positive-area overlap with a corner of the grey face. These three overlaps can be made arbitrarily small without allowing the white cubes

Figure 50
Step 2 in the 24-cube arrangement

P and Q to project to the left beyond the vertical line CD. As a result angle e and distance d can also be made arbitrarily small.

This pattern of seven cubes goes on the front and rear faces of the ¡grey cube. Then one cube goes exactly on top of the grey cube, another goes flush against the grey cube's base and two more cubes abut the right face of the grey cube. Although 18 cubes now abut five faces of the grey one, its sixth face (on the left) remains completely exposed. Figure 51 shows the grey cube with the exposed face toward you. On both its left and right sidies are seven cubes; they are not shown in the drawing. (Also not shown are the single cubes above and below and the two cubes that abut the grey cube's back face.) Cube K is placed so that it overlaps the top of the grey face along a thin horizontal strip of height d that can be arbitrarily small. This allows five more cubes to abut the face, below cube K, as shown, bringing the total number of touching cubes to 24 (7 + 7 + 1 + 1 + 2 + 1+5).

Figure 51

Figure 51

A Full proof of this construction would be long and tedious, but interested readers should have no difficulty convincing themselves that it can be done in spite of the extremely minute overlaps that are involved. The 24-cube solution is probably maximum, although proving it appears to be formidable. Until that is done there will remain the gnawing suspicion that one or more additional white cubes can somehow be squeezed in.

Theodore Katsanis posed an interesting related problem. What is the minimum number of unit cubes that can touch another unit cube in such a way that'no other cube can be added? If "touch" is defined as in the original problem, the answer is obviously six. Suppose, however, we enlarge the meaning of touch to include contact along edges and corners. The maximum problem is again trivial, answered by 26 cubes, but the minimum problem is not. The lowest Katsanis could get is nine, but perhaps some reader can do better.

(34) Commented P. H. Lyons: "I hope some readers tried other languages, such as Hawaiian. If the letters of the alphabet need not be in alphabetical order, I have a fair-sized list of other answers in English."

(35) William B. Friedman proposed placing the rope so that half of it is high above the water and the other half still higher. The man could then walk along the lower rope, holding on to the upper one, and not even get wet. If the rope is hemp, wrote P. H. Lyons, the man could smoke it and fly to the island.

(36) This question (about the boy, the girl and the dog) stirred up a hornet's nest. Some mathematicians defended the answer as being valid, others insisted the problem has no answer because it is logically contradictory. There is no way the three can start moving, it was argued, because the instant they do the dog will no longer be between the boy and the girl. This plunges us into deep waters off the coast of Zeno. The issue is discussed in detail in Chapter 13.

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