Old challenge (Jonathan Kravis and Dave McMath)
A process involving infinitely many ping-pong balls is carried out in one hour as follows. In the first half hour, two balls are placed into a huge empty barrel and then one is removed and discarded. In the next 15 minutes, two more balls are put in and then one of the three inside is removed. At each step, in half of the remaining time, two more balls are added and one is removed. How many balls are in the barrel at the end of the hour after infinitely many steps?
After n steps, there are n balls in the barrel. After infinitely many steps, there might be infinitely many in the barrel or there might be none at all or any number in between!
Suppose for example that at the nth step you remove the nth ball. Then every ball gets removed at some time, and none are left. Of course if at the nth step you remove ball 2n, then balls 1, 3, 5, ... will never be removed. Infinite processes often defy intuition.
If you remove balls at random, it turns out that with 100 percent probability there will be no balls left. For example, the probability that ball 1 remains after one step is 1/2, after two steps is (1/2)(2/3) = 1/3, after three steps is (1/2)(2/3)(3/4) = 1/4, after n steps is 1/(n+1), and after infinitely many steps is 0.
According to Jim Henry, for an engineer, there will be only finitely many balls in the barrel, because after the 19th step after 59 minutes and 59.99 seconds, the engineer will say the hour is up.
Likewise for a quantum-relativist physicist, there will be only finitely many balls in the barrel, because the later ones would have to surpass the speed of light to get inside.
More on arithmetic modulo 7 (Nick Drozdoff)
"I am a high school physics teacher, ex-electrical engineer, and a current physics graduate student. [In the June 27 discussion of arithmetic modulo 7] were many equations such as 52 = 25 = 4. This seems a peculiar form of mathematics. I don't personally see how 25 = 4. What I can see is that 25 divided by 7 is three with a REMAINDER of four, but that is not what was written. What was written was 25 = 4.
"High school kids do this in calculations all the time. They string their work together with equal signs skipping steps in favor of some sort of shorthand. It drives me nuts because they have difficulty checking their work later. I encourage them to write each step of a calculation out. I discourage the tendency to string things together with equal signs that lead to visual/mathematical absurdities.
"Is this a common practice with math instruction? If so, I am appalled and shocked. I ask because this problem is not limited to any particular school. It's everywhere. I am concerned when I see this sort of thing in print in a place where people might regard such techniques as being accepted practice with experts, such as in The Christian Science Monitor, a respected newspaper not known for egregious mistakes. Please let me in on what this practice is all about.
"I do want you to know that I appreciate your column. How many newspapers have the guts to publish something that actually is interesting and requires or, rather, invites thought!"
Math Chat replies: All mathematics teachers know exactly what Mr. Drozdoff is talking about. Students who mean "3x = 6, therefore x = 2" instead incorrectly write "3x = 6 = x = 2." In contrast, we were not just stringing together different steps, but using the notion of "equality modulo 7" consistently throughout for numbers with the same remainder after dividing by 7, such as 25 and 4. To emphasize that this is a new kind of equivalence, mathematicians sometimes use an equal sign with three bars and call it congruence instead of equality. On the other hand, mathematics needs to use many new systems with their own numbers and identities, and, for example, 25 and 4 could be considered different symbols for the same numbers (or "congruence classes") in a new arithmetic modulo 7, where 25 really does equal 4.
Last weekend 850 mathematicians and students gathered in Atlanta for the annual Mathematical Association of America Mathfest, including a celebration of the mathematics of the late Paul Erds and a mathematical play by my Williams colleagues Colin Adams and Edward Burger.
A team of six American high school students tied for fourth place at the 38th International Mathematical Olympiad in Mar del Plata, Argentina, July 18-31. The top five teams out of 82 were China, Hungary, Iran, the United States, and Russia. On the US team were: Carl Bosley, Washburn Rural High School, Topeka, Kan.; Nathan Curtis, Thomas Jefferson High School for Science and Technology, Alexandria, Va.; Li-Chung Chen, Monta Vista High School, Cupertino, Calif.; John Clyde, New Plymouth (Idaho) High School; Josh Nichols-Barrer, Newton South High School, Newton Center, Mass.; and Daniel Stronger, Stuyvesant High School, New York.
What is the chance that two people will randomly choose the same letter of the alphabet at random? What is the chance that the names of two random US presidents begin with the same letter? Richard Thorne noticed some common first letters in the winners of June 27: Bright, Chang, Chapman, Daniels, Dravecky, Hildebrand, Howard, Longworth, Randolph, Rice. Estimate the chance that at least two of the names of 10 random Americans will start with the same letter.
* To be eligible for 'Flatland' and other book awards, send answers and new questions to:
Bronfman Science Center Williams College,
Williamstown, MA 01267
or by e-mail to: