536 Puzzles and Curious Problems

(Elliott) #1
384 Answers

16,23,25, and 31 inches from one end. Another solution is I, I, I, 1,6,6,6,6,5.
This puzzle may be solved in, at fewest, sixteen different ways. I have
sought a rule for determining the fewest possible marks for any number of
inches, and for at once writing out a solution, but a general law governing all
the multiplicity of answers has still to be found.
[Although no general rule has yet been found for the ruler problem, con-
siderable progress has been made since it was proposed by Dudeney. The
reader is referred to John Leech's paper, "On the Representation of I, 2, ... ,
n by Differences," in the Journal of the London Mathematical Society, Vol. 31,
1956, pp. 160-169. Leech discovered that eight marks are also sufficient for
marking a 36-inch yardstick so that all integral values from I to 36 can
be measured. The reader may like to search for this pattern. Had Dudeney
known it, he would have surely found it preferable to his broken yardstick of
33 inches.-M. G.]


  1. THE SIX COTfAGES


If the distances between the cottages are as follows, in the order given, any
distance from one mile up to twenty-six inclusive may be found as from one
cottage to another: I, 1,4,4,3, 14 miles round the circular road.
[This problem is, of course, a circular version of the preceding one. As be-
fore, Dudeney could have increased the length of the "measuring stick" (in
this case, the road) without altering the other aspects of his puzzle. It turns
out that six cottages can be placed on a circular road of 31 miles so that all
integral distances from 1 to 30 are represented by a distance, on the circle,
between a pair of houses. It is not hard to see that for n houses the maximum
number of different ways to measure a length between houses is n(n - I).
For n = 6, the formula gives 30, so in this case it is possible to place the 6
houses on the 3 I-mile road so that no distance between any pair of houses
duplicates any other distance. Similar optimum solutions can be found when
n = 1, 2, 3, 4, or 5. See Michael Goldberg's solution to Problem EI76 in
American Mathematical Monthly, September 1965, p. 786.-M. G.]


  1. FOUR IN LINE


There are nine fundamentally different arrangements, as shown in the illus-
tration, the first, A, being the arrangement given as an example. Of these, D,
E, and I each give eight solutions, counting reversals and reflections as ex-

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