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More Mathematical Morsels

Mathematical Association of America
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As the author says in his preface, the book “might well be called Gleanings from Crux Mathematicorum.” Crux Mathematicorum was founded in Canada in 1975 by Léo Sauvé and Frederick G. B. Maskell as a problem journal for undergraduate students of mathematics. It was originally called Eureka but those in charge of the journal of the same name published by the Archimedean Society of the University of Cambridge took notice (after some time) and its title was changed in 1978. It was later taken over by the Canadian Mathematical Society, merged with another problem journal in 1997, and continues under the unfortunate name of Crux Mathematicorum with Mathematical Mayhem. It contains short articles and book reviews but it is mostly devoted to problems and solutions, those provided by readers and also those that appear in various Olympiad-like competitions throughout the world. The problems are at the high school and undergraduate level though, as usual, most of the solvers are professional mathematicians.

Honsberger takes fifty-seven problems, almost all from Crux, and provides his own clear and leisurely solutions. He selected only elementary problems with, he says, the aim of providing enjoyment. Murray Klamkin had an Olympiad Corner column in Crux from 1979 to 1986 and problems and solutions from it are included as well.

I suppose there are members of the Mathematical Association of America who are immune to the appeal of problems and solutions, but I hope their number is not large. For the rest of us, the book contains delights. You probably do not know the possible values of the greatest common divisor of n2 + 100 and (n + 1)2 + 100. One value is 1 and as you checked n = 1, 2, …, 199 and found no other you might be tempted to conclude that there is none. However, every now and then, the gcd 401 appears, for the first time at n = 200. The proof, after you see it, is obvious, but probably would not have occurred to you right away.

Morsel 45 asks which is the correct answer to a multiple-choice problem:

  1. All of the below.
  2. None of the below.
  3. All of the above.
  4. One of the above.
  5. None of the above.
  6. None of the above.

It doesn’t matter what the problem is, the answer is (e).

To show that, for a, b, c, and d in [0, 1],

(1 – a)(1 – b)(1 – c)(1 – d) + a + b + c + d ≥ 1

you do not use the arithmetic-mean geometric-mean inequality but rather a method that applies no matter how many variables there are.

Mathematics is inexhaustible, and so is elementary mathematics. If these sorts of things appeal to you — how could they not? — this book is well worth having.

Woody Dudley used to be able to solve problems, even some that appeared in the Monthly, but, now that he’s retired, has decided it’s enough to read solutions. 




Date Received: 
Friday, January 27, 2006
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Ross Honsberger
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Underwood Dudley
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Friday, January 14, 2011