Mathematical puzzles for your intellectual amusement.
March 31 Problem. What is the largest architectural example of a regular polygon on the campus of Goshen College?
Mark Miller’s Solution. The structure of the Marine Biology research facility in Florida (part of Goshen’s campus) is square.
Penina Christine Acayo’s Solution. The best example of a regular polygon on campus is the College Mennonite Church. [Look at the walls and windows of the chapel around the upper level to see the 48-sided regular polygon.]
Nick Hans Bouwman’s Solution. If the tennis courts are considered architectural, they win with a size of 72 by 72 meters = 5184 meters squared. If architectural refers to buildings only, the square-like structure on top of the church-chapel is about 53 by 53= 2809 meters squared.
March 23 Problem. A digital watch displays hours and minutes with am and pm. What is the largest possible sum of the digits in the display?
Solution by Dan Stutzman. if we’re limited to summing digits, i believe 9:59, 9+5+9=23 is the largest sum. unless we can speak hex and take 9:59am, ignore the ‘m’ and add 9+5+9+a = 0×21, expressed in decimal as 33.
March 10 Problem. Let A denote multiplication, which is just repeated addition. So, 2 A 3 = 2 * 3 = (2 + 2) + 2 = 6. Let B denote repeated A. So, 2 B 3 = (2 A 2) A 2 = (2 * 2) * 2 = 8. Let C denote repeated B. So, 2 C 3 = (2 B 2) B 2 = (2 A 2) B 2 = (2 * 2) B 2 = 4 B 2 = 4 A 4 = 4 * 4 = 16. Let D denote repeated C, E denote repeated D, and so forth. What is 2 D 3? What is 2 E 3? What is 2 Z 3?
Solution by Seth Unruh. 2d3=(2a2)c2=4c2=4b4=4^4=256 and
2e3=(2a2)d2=4d2=4c4=4b4b4b4=(256b4)b4=256^16=2^128 = 340,282,366,920,938,463,463,374,607,431,768,211,456
March 3 Problem. Arcadia University advertises a 120 percent undergraduate participation in study abroad (see http://tinyurl.com/ahjtmy). The prize this week will go to the best explanation for how this is possible and/or for the
best suggestion for how Goshen College can top Arcadia.
Solution by Judy Weaver. Arcadia University achieves a 120% study abroad rate by sending some students abroad more than once during their four years of college. They require study abroad, as GC does, by offer several majors that require a year abroad, as GC does, and they do two things that GC doesn’t: they take first years students for a semester abroad and they ship masses of students to world capitals during spring break. Goshen can top this easily! We already send 80-85% of our students abroad through SST and year abroad programs. Out of a student body that is around 1,000, we only need to send another 200 students abroad to reach 100% – we can do this by making sure that all sports teams include a stop in Canada whenever they travel north, crossing at the nearest point (probably Detroit? – at 3 1/2 hours away) – they can go across the border for a sandwich, write a paper about it and receive credit. To match Arcadia’s 120%, we develop a relationship with the Windsor, Ontario Mennonite Fellowship (just across the border from Detroit), and take the entire first-year class of about 200 students on a service assignment there over spring break (they get credit for that, too, of course). Then to top Arcadia, we send all the choirs for weekend tours to Niagara Falls (Canadian side) where they practice singing over the roar of the falls and learn to stay well hydrated, which is important for singers. Even allowing for some people who are unable to participate, this will put us at about 130%. We could also have all these travelers cross back in the U.S. and back again into Canada before returning home, count it as two trips abroad and double our numbers, but that would be unethical.
Solution by Jim Miller. I assume that they’re counting people who go abroad more than once. e.g., if 60% of their students study overseas one time, and 30% study overseas twice, the total is 60% + 2 x 20% = 120%. We might be able to get our number up to 100% or more if we counted that way. You can see similar kinds of number in risk factors for disease. Research might show, for example, that 30% of heart attacks are caused by hypertension; 50% are caused by obesity; 50% are caused by hyperlipidemia; 40% are cause by coach potato-itis (lack of exercise); 40% are caused by hereditary factors. You can very quickly get totals that add up to more than 200%. It just means that most people with coronary heart disease have more than one risk factor, and that if any one of them had been eliminated, they would have avoided the disease. This is a gross oversimplification, of course, but the basic concept is true.
Solution by Stan Miller. Paul Harvey used to report on “the rest of the story”.
Arcadia’s claim comes from a report published by Open Doors of the Institute of International Education. They derive these particular statistics by “dividing the total number of undergraduates who studied abroad in a given year (as reported in the Study Abroad survey), by the total number of undergraduate completions (from IPEDS). Due to various factors, such as students studying abroad more than once, students dropping out before graduation and differing cohort sizes from year to year, participation rates may exceed 100%.”
This is one of the most blatant misuses of statistics that I’ve ever seen. I’m surprised they can get away with it in the higher education community. I complete their annual survey so I challenged their methodology a few years ago. They defended it by saying the participation rate was only meant to be an “estimate”.
February 2 Problem. I had a rough time last week and so I have fallen behind in posting these problems. It got me to thinking. If I experience ten random bad events per year, what is the probability that I will experience more than one bad event in a single week?
Solution by David Housman. Assume that there are exactly 52 weeks in a year and that each bad event is equally likely to fall in any week. Then the probability that no bad event occurs in a particular week is (51/52)10 = 0.8235, the probability that exactly one bad event occurs in a particular week is 10(1/52)(51/52)9 = 0.1615, and the probability that more than one bad event occurs in a particular week is 1 – 0.8235 – 0.1615 = 0.0150.
January 23 Problem. Is Barack Obama a maraca kabob? Well, “maraca kabob” is an anagram of “barack obama.” This motivates this week’s problem. How many different ways can the letters “barackobama” be rearranged?
Solution by Chaim Hodges. Under the assumption there are no spaces and each combination does not have
to make a word the answer comes out to be (11factorial/4factorial) /
2factorial. The reasoning for this is that there are 4 a’s and 2b’s and
that cuts out a significant portion of word combinations that can be made.
11factorial = 39,916,800
4 factorial = 24
2 factorial = 2
(39,916,800 / 24) / 2 = X
39,916800 / 24 = 1,663,200
1,663,200 / 2 = 831,600
The number of ways that barackobama can be made without repeating is
January 15 Problem. Let’s start the semester with a game! You simply submit a number between 0 and 100. The person who submits the number closest to 70% of the median of the numbers submitted will win a prize (ties will be broken randomly). For example, if a total of five people submit the numbers 13.2, 50, 60, 73.4286, and 94, then 70% of the median is (0.7)(60) = 42, and the person who submitted 50 would win the prize. You may also submit an explanation for your number choice. The person who submits the best explanation will win a prize, too. Go to http://tiny.cc/ES57Y to submit your number and explanation.
Solutions by Lisa Goldsmith and Bob Toews. Thirteen numbers were submitted, and 70% of the median turned out to be 21.7. Lisa Goldsmith chose 23 “[b]ecause I’m psychic, so I know what you are thinking.” Perhaps she is and does because Lisa chose the closest number!. Bob Toews had the best explanation: “Not being able to predict what number someone will chose, I’m assuming some people will confuse mean with median and hope that they determine the mean of the numbers chosen will be around 50 and will multiply that by 0.7 (35). I’m guessing the answer is 70% of that number or 24.5. :-)”