MAY 18, 2000
One the Most Common Digit? Neville Barnard (and similarly Dino Surendran, of the University of Zimbabwe) write: "As a non-mathematician I was intrigued to read of Benford's Law, which states that in a large group of numbers, such as accounting figures or census data, the distribution of initial digits [favors lower digits]: "1" 30% of the time, "2" 18% of the time," down to "9" 5% of the time.
Math Chat responds: Lower digits are more common as initial digits simply because they come first. Half of the numbers from 1 to 20 begin with a 1.
To predict the expected percentages, assume that the occurrence of numbers is inversely proportional to their size, which means that the logs (the fractional powers of 10 that give you the numbers) are uniformly distributed. Then the probability of an initial 1 is log 2 - log 1 or about 30% and so on:
|1||log 2 - log 1 = 30%|
|2||log 3 - log 2 = 18%|
|3||log 4 - log 3 = 12%|
|4||log 5 - log 4 = 10%|
|5||log 6 - log 5 = 8%|
|6||log 7 - log 6 = 7%|
|7||log 8 - log 7 = 6%|
|8||log 9 - log 8 = 5%|
|9||log 10-log 9 = 5%|
How to Ace Calculus. If you know someone taking calculus, there are a couple of books which (s)he might find a real boon. How to Ace Calculus, the Streetwise Guide, by Colin Adams, Joel Hass, and Abby Thompson, provides an irresistibly humorous guide to calculus and studenthood:
page 1: "If you are reading this introduction then this book is probably not for you."
page 10: "General Principles of Acing Calculus: 1. Buy this book. 2. Pick the right professor. 3. Pay attention. 4. Do the homework. . . . 10. Avoid the dark side."
Available at the MAA bookstore at this website. Also see http://www.geocities.com/CollegePark/Bookstore/2072/ace.html.
Misteaks by Barry Cipra shows a calculus student how to catch mistakes by looking for obvious ones and learning how to make rough estimates:
"A man pulling a rope in a related rates problem will not in general pull at velocities exceeding the speed of light. Nor will a rope 500 feet long ever enclose an area this size of New Mexico. And it certainly won't enclose a region whose area is negative."
The great thing is that if you learn these "cheap tricks" you understand calculus much better. Recently republished by A K Peters (http://www.akpeters.com).
Old Challenge (Al Zimmermann). On ABC TV's "Who Wants to be a Millionaire," what should your strategy be if the $500,000 question is: "With how much money will you leave here tonight:
(A) $32,000 (B) $250,000 (C) $500,000 (D) $1,000,000"?
Answer. Math Chat thinks that the wisest course is to walk away with the $250,000 already won, since all four answers are fraught with grave danger. If you answer (A), even if you're right, that must mean you'll eventually make a mistake and end up with $32,000; and it seems as if you cannot be wrong, because then you'd leave with $32,000 and you'd be right! If you answer (B), (C), or (D), the show could just say that you're wrong and send you home with $32,000 (and you would indeed have been wrong). In particular, once you answer (B), $250,000 is no longer a possibility, and you have to be wrong.
Eric Brahinsky points out that with a kind host, you might just make out well with (C) or (D). After all, if you pick (C), the host could say "right" or "provisionally right" and let you walk with $500,000. (If you went on to the $1,000,000 question, no matter what the outcome, your previous answer would become retroactively "wrong," and you'd leave with just $32,000.) If you pick (D), again the host could say "provisionally right" and let you try to go on to win $1,000,000. Of course if you got the $1,000,000 question wrong, your previous answer would become retroactively wrong, and you'd leave with just $32,000. So to pick (D) you'd have to be rather confident of both the host's kindness and your own ability.
Incidentally, Walter Wright reports the following question from the quiz show "Greed":
Which of the following does not happen once every four years:
a) Summer Olympic Games
b) World Cup Football
c) Nobel Prize Awards
d) U.S. Presidential Elections
e) Leap Year
Wright remarks: "Now, if I had been playing, I would have been about 99.8 percent sure that the Nobel prizes are awarded every year, and so I would have been extremely confident that "c" was the "right" answer. But I would have been 99.999999 percent sure that e) is the right answer, since we skip leap years occasionally (when the year is evenly divisible by 100, but not by 400, such as 1900).
So should I have answered "e," knowing that I could have challenged the game show and proven that "e" is a correct answer, or should I have answered "c," being extremely confident (but not certain) that this is correct and also confident that it was the answer they were looking for?
Of course, a lot depends on knowing what the game show would do in this case: maybe if I answered "e" and proved my case, the show would invalidate the question and replace it with another question (that undoubtedly would be much harder!)."
New Challenge. Consider the sequence of integers 10n + 1:
11, 101, 1001, . . . .
Are there infinitely many primes? Infinitely many composite numbers?
Send answers, comments, and new questions by email to Frank.Morgan@williams.edu, to be eligible for Flatland and other book awards. Winning answers will appear in the next Math Chat. Math Chat appears on the first and third Thursdays of each month. Prof. Morgan's homepage is at www.williams.edu/Mathematics/fmorgan.
Copyright 2000, Frank Morgan.