My Best Mathematical and Logic Puzzles by Martin Gardner

Author:Martin Gardner
Language: eng
Format: epub, pdf
Publisher: Dover Publications
Published: 2013-03-15T16:00:00+00:00

16. At any given instant the four bugs form the corners of a square which shrinks and rotates as the bugs move closer together. The path of each pursuer will therefore at all times be perpendicular to the path of the pursued. This tells us that as A, for example, approaches B, there is no component in B’s motion which carries B toward or away from A. Consequently A will capture B in the same time that it would take if B had remained stationary. The length of each spiral path will be the same as the side of the square: 10 inches.

If three bugs start from the corners of an equilateral triangle, each bug’s motion will have a component of 1/2 (the cosine of a 60-degree angle is 1/2) its velocity that will carry it toward its pursuer. Two bugs will therefore have a mutual approach speed of 3/2 velocity. The bugs meet at the center of the triangle after a time interval equal to twice the side of the triangle divided by three times the velocity, each tracing a path that is 2/3 the length of the triangle’s side.

For a generalization of this problem to n bugs at the corners of n-sided polygons see Chapter 24 of my Sixth Book of Mathematical Games from Scientific American (W. H. Freeman, 1971).

17. When Jones began to work on the professor’s problem he knew that each of the four families had a different number of children, and that the total number was less than 18. He further knew that the product of the four numbers gave the professor’s house number. Therefore his obvious first step was to factor the house number into four different numbers which together would total less than 18. If there had been only one way to do this, he would have immediately solved the problem. Since he could not solve it without further information, we conclude that there must have been more than one way of factoring the house number.

Our next step is to write down all possible combinations of four different numbers which total less than 18, and obtain the products of each group. We find that there are many cases where more than one combination gives the same product. How do we decide which product is the house number?

The clue lies in the fact that Jones asked if there was more than one child in the smallest family. This question is meaningful only if the house number is 120, which can be factored as 1 × 3 × 5 × 8, 1 × 4 × 5 × 6, or 2 × 3 × 4 × 5. Had Smith answered “No,” the problem would remain unsolved. Since Jones did solve it, we know the answer was “Yes.” The families therefore contained 2, 3, 4 and 5 children.

This problem was originated by Lester R. Ford and published in the American Mathematical Monthly, March 1948, as Problem E776.

18. The heads of the twiddled bolts move neither inward nor outward.

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