Today's column (No. 26) marks the first full year of Math Chat. Many thanks to readers, in communication or silent, for their willingness to have some fun thinking about mathematics. The winning answers mentioned today are just part of the large reader response.

Old speed limit challenge (Marc Abel)

When a state increases a highway speed limit from 55 to 65 miles per hour, by what percent does the road capacity (in cars per hour) change?

Answer

Jean Hunter says it best: If traffic moved at the same spacing at both speeds, then the capacity would increase by the same amount, 10/55 or about 18 percent. However, if the traffic spacing is governed by the safety rule of one car-length between cars for every 10 m.p.h. of speed, then capacity increases only about 2.5 percent.

According to Glenn Grigg, a traffic engineer, "the Highway Capacity Manual states that the ideal capacities of 2,200 and 2,300 vehicles per lane per hour can be sustained over a broad range of speeds." Scott MacCalden reports that "for freeway driving during commute periods (when almost all vehicles are passenger cars, and the drivers are familiar with the road) average time intervals or 'headways' between cars may vary between 2.0 and 1.5 seconds. This is much more than on the Indy 500 track where headways can be considerably less than 1 second and average speeds reach 200 m.p.h.!" His policy recommendation is to eliminate bottlenecks and encourage carpools and buses.

Jan Smit observes that at high speeds where safe braking distances become more important than reaction time, road capacity might decrease as speeds increase.

Four 4s and large primes

A furious contest has been raging in response to William Foster's challenge to readers of May 22 to represent with four 4s a larger prime than 257 = 44+4/4. James Grimm and Michael Stern proposed 577 = 4!x4! + 4/4. Garrett Gray and Michael Eastep tendered 331,777 = 4!4 + 4/4. Eric Brahinsky proposed 479,001,599 = 12! - 1 = (4!/4)! - 4/4, which he found in a list of primes in the Penguin Dictionary of Curious and Interesting Numbers by David Wells. Foster countered with 1,197,503,999 = 12!/.4 - 1 = ((4!/4)! - .4)/.4, which he confirmed prime with his own computer program, with the fateful words, "I doubt that a larger one will be found by publishing date." The next day he found on the Net an algorithm for testing primes and verified that the following champion is prime:

16723757283622817643970295135225931466881865832845074525739178380754083893112407473417185413179962453270394094733035226381336021473520304649746202785974125014912401303142399999999999999999999999999999 = (5!!-.4)/.4 = ((4/.4)!!-.4)/.4.

This "Elliptic Curves and Primality Proving" algorithm of Atkin and Morain (1993) tests for primes with up to 1000 digits. If you do not insist on logical 100 percent certainty, there are much faster probabilistic tests to verify primes with 99.99 percent or any desired level of certainty. Any one test may let certain "pseudo-primes" slip through. Prof. Andrew Granville of the University of Georgia reported that in a major advance Jon Grantham's recent PhD thesis showed that a dozen or so supposedly different kinds of pseudo-primes are really all examples of a single concept based on the so-called "Frobenius endomorphism."

Meanwhile Marc Abel tried to represent the largest known (Mersenne) primes (see Math Chat Sept. 13 and 27, 1996 and Jan. 17, 1997) by four 4s and then wrote a computer program to exhaustively find all numbers representable by four 4s. The program said that our old stumper of 73 is indeed impossible, as well as 87, 93, 99, 103, 105, and many others, eventually including most odd numbers and then more and more even numbers. The last few odd possibilities on his list through 3,628,799 unfortunately are not prime.

Japanese entrance exam

The Japanese entrance exam question, which ran with an article on Wednesday, May 21, 1997, may have stumped the Monitor staff, but not our readers. Cecil Rousseau, Jeffrey Northridge (see their letters on Page 20), Diana Palenz, John Sullivan, Marc Abel, and Misha Chkhenkeli solved it. They note that "rectangle" should instead read "tetrahedron" (triangular pyramid), as in the Japanese original (literally, "four-faced body"). Reiko Yamada, associate professor of Japanese at Williams College, verifies that the problem provided by the American Federation of Teachers was mistranslated.

The unit sphere must be centered at (a/2, 1/2, b/2) and thence a2 + b2 = 3. The radius r of a sphere inscribed in a tetrahedron of surface area A and volume V satisfies the formula r = 3V/A, which becomes 1/r - 1/a - 1/b = (1+a2)/a + (1+b2)/b.

The right-hand side is smallest when a and b are equal and hence both equal to (3/2) (because a2 + b2 = 3), making the right-hand side (20/3) and solving the first half of the problem. Similarly the same values yield the smallest value of 1/r = 2(2/3) + (20/3) = 2(2/3) + 2 (5/3), solving the second half of the problem.

New challenge (Howard Sheldon)

What is the remainder when you divide 5999,000 by 7?

* Send answers, comments, and new questions to: Math Chat, Bronfman Science Center, Williams College, Williamstown, MA 01267or by e-mail to: Frank.Morgan@williams.edu

The best submissions will receive 'Flatland' and other book awards.