V5 x2

Solve for x and change the result to the nearest integer. It is 618,034. Because no other integer, when related to 1,000,000, gives a closer approximation of the golden ratio, 618,034 is the next-to-last term of the longest possible chain of positive integers in a generalized Fibonacci series ending in 1,000,000. One can now easily work backward along the chain to the first two terms. (This method is explained in Litton Industries' Problematical Recreations, edited by Angela Dunn, Booklet 10, Problem 41.)

6. Contrary to most people's intuition, your best bet is that the top card is a black ace.

The situation can be grasped easily by considering simpler cases. In a packet of three cards, including the two black aces and, say, a king, there are three equally probable orderings: AAK, AKA, KAA. It is obvious that the probability of the first ace's being on top is 2/3 as against 1/3 that it is the second card. For a full deck of 52 cards the probability of the top card's being the first black ace is 51/1,326, the probability that the first black ace is second is 50/1,326, that it is third is 49/1,326, and so on down to a probability of 1/1,326 that it is the 51st card. (It cannot, of course, be the last card.)

In general, in a deck of n cards (n being equal to or greater than 2) the probability that the first of two black aces is on top is n — 1 over the sum of the integers from 1 through n — 1. The probability that the first black ace is on top in a packet of four cards, for instance, is 1/2.

The problem is given by A. E. Lawrence in "Playing with Probability," in The Mathematical Gazette, Vol. 53, December 1969, pages 347—354. As David L. Silverman has noticed, by symmetry the most likely position for the second black ace is on the bottom. The probability for each position of the second black ace decreases through the same values as before but in reverse order from the last card (51/1,326) to the second from the top (1/1,326).

Several readers pointed out that the problem of the first black ace is a special case of a problem discussed in Probability with Statistical Applications, by Frederick Mosteller, Robert E. K. Rourke and George B. Thomas, Jr. (Addison-Wesley, 1961). Mosteller likes to call it the "needle in the haystack" problem and give it in the practical form of a manufacturer who has, say, four high-precision widgets randomly mixed in his stock with 200 low-precision ones. An order comes for one high-precision widget. Is it cheaper to search his stock or to tool up and make a new one? His decision depends on how likely he is to find one near the beginning of a search. In the case of the 52-card deck there is a better-than-even chance that an ace will be among the first nine cards at the top of a shuffled deck or— what amounts to the same thing—among the first nine cards picked at random without replacement.

7. The three essentially distinct solutions of the dodecahe-dron-quintomino puzzle are shown on Schlegel diagrams in Figure 17. They were first published by John Horton Conway, the inventor of the problem, in the British mathematical journal Eureka for October, 1959, page 22. Each solution has a mirror reflection, of course, and colors can be interchanged without altering the basic pattern. The letters correspond to those assigned previously to the 12 quintominoes. The letter outside each diagram denotes the quintomino on the solid's back face, represented by the diagram's perimeter.

Three basic solutions to the dodecahedron-quintomino problem

Conway found empirically that whenever the edges of 11 faces were correctly labeled, the 12th face was automatically labeled to correspond with the remaining quintomino. He did not prove that this must always be true.

Because the regular dodecahedron is the "dual" of the regular icosahedron, the problem is equivalent to coloring the edges of the regular icosahedron so that at its 12 vertexes the color permutations correspond to the permutations of colors on the 12 quintominoes.

In 1972 a white plastic version of Conway's puzzle was on sale in the United States under the name "Enigma." Patterns of black dots were used instead of colors.

8. The scrambled quotation is "There is no frigate like a book/To take us lands away." It is the first two lines of a poem by Emily Dickinson.

9. There is certain to be at least one column of blank spaces on that typed page. Assume that the sentence is n spaces long, including the first space following the final period. This chain of n spaces will begin each typed line, although the chain may be cyclically permuted, beginning with different words in different lines. Consequently the first n spaces of every line will be followed by a blank space.

10. Only the following eight triplets have a product of 36: 1,

1, 36; 1, 2, 18; 1, 3, 12; 1, 4, 9; 1, 6, 6; 2, 2, 9; 2, 3, 6, and 3, 4, 3. Speaker A certainly knew his own house number. He would therefore be able to guess the correct triplet when he was told it had a sum equal to his house number—unless the sum was 13, because only two triplets have identical sums, 1 + 6 + 6 and 2 + 2 + 9, both of which equal 13. As soon as A was told that B had an oldest child he eliminated 1, 6, 6, leaving 2,

2, 9 as the ages of B's three children.

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