In a technical, legalistic sense, the semester started last week. But as far as I'm concerned, the semester doesn't really begin until Problem of the Week returns!
Our theme for this semester: Clock Problems.
That's not code for modular arithmetic or anything. I mean it literally. Every problem this term will feature clocks in some way. Some are fairly easy, some are a bit harder, though I wouldn't say any of them are killer. Keep in mind that I deliberately keep the problems a little on the basic side, since I want students in the lower level math classes to be able to participate. So, nothing that requires anything more than algebra or geometry, and maybe some trigonometry.
You also do not need to worry that students will look on this blog and see the solutions put forward in the comments. Participation is entirely optional, and they are not receiving any sort of course credit for handing in a solution. The kind of student who enjoys participating in this is also the kind who will not be looking for a shortcut to the solution.
So feel free to have at it in the comments!
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I guess the first step is to calculate that the sum of all numbers 1-12 is 78. 78 doesn't divide evenly by 4, so it cannot possibly the case that our lines will cross - we will end up with 3 sections, and the sum of the numbers in each must be 26.
With a little trial and error, draw a line between the 10 and 11 to between the 2 and the 3, and a second line between 8 and 9 and between 4 and 5. Then we have:
1 + 2 + 11 + 12 = 26
3 + 4 + 9 + 10 = 26
5 + 6 + 7 + 8 = 26
I got the same answer, using basically the same logic; (1) [(N+1)(N)/2]/3 = 26, (2) hey, 11+12+1+2 is 26, (3) find a second group of 26 in the remainder.
Yes, this one is easy. 12*13/2 = 78 which is not a multiple of 4, so you are looking for three sections each summing to 26. Then you notice that you can draw a line such that the sum of the number on one side and the corresponding number on the other side is 13, so each of your sections must have two such pairs. Thus 11-12-1-2 form one section and 5-6-7-8 form a second section, leaving 3-4-9-10 for the third.
As for the joke about the clock striking 13: 1984 opens with that line. It makes sense in contexts where you are using a 24-hour clock (e.g., if you are using UTC, as I often do in my work, you always use a 24-hour clock). At the same time (pun not intended), the number 13 is associated in Western culture with bad luck, so you get an impression that something is off.
As for the joke about the clock striking 13...
Aside on the cultural references; my very first thought when reading the POTW, before even getting to the problem, was that Jason's students aren't going to get the music references. The clock problem is probably easier for them to figure out than the answer to "who are the Bangles?" :)
I think this was probably not the intended solution, but I believe it's a valid solution to the problem as stated. Draw a line from the center of the clock to a point between the 9 and 10. Draw a second line from the center to a point between the 3 and 4. You have divided the clock into two sections each with numbers totalling 39.
I hope you do get to some mod 12 puzzles though. I'm totally ready for them because they are rife in music theory (this week's problem has bearing on musical inversion - that is themes that mirror such that when one goes up the other goes down, and by the same interval).
I have a lovely doozy that I'll be glad to post at some point. No time now - gotta finish a piano concerto!
Yes, Jason's dating himself with that one. Most of his students were born 5-10 years after the Bangles' cover of the song in question was a hit. And I imagine that many of them won't have a clue who Art Garfunkel is. They may have heard of Paul Simon, who has been putting out an album every five years or so since Graceland (including one released this year), but Simon's songwriting style has been out of fashion among the 15-25 crowd for some time now.
Sean@5: A line in geometry is assumed to extend infinitely in both directions, so your solution is not allowed. You will get two regions that sum to 39 and two that sum to 0. The case where your two lines coincide is considered to be not sporting fair.
One possible way to get around Eric Lund's objection to Sean T's solution is to draw a line from between 9 and 10 to between 3 and 4. Next, draw another line exactly on top of the first line. This will result in two sections of 39 each.
While technically true, Eric, I actually think that Sean T's solution is sufficiently clever that if I were Jason I'd still give full credit for that. :)
@Gazza: The problem with Sean T's solution is that it begins at the center of the circle. But we have no point at the center of the circle to begin with. We only have a clock with numbers, but no center. I don't see a center point from which to begin drawing in the example provided by Jason.
Oops, I should have said "provided by Sean T" not Jason.
My solution simply requires two lines drawn on top of each other. Is that forbidden in geometry? Will one line weigh down on the other? Will there be a slight gap between those two lines? Well, since lines don't really have two dimensions in geometry, I don't see what the problem is. Sure, in the physical world drawing such a thing would be impossible, but pure geometry is often viewed as abstract science, as it should be. No one can draw a straight line.
The Bangles, and Simon and Garfunkel, are entertainers, so there is at least some chance the students will have heard of them. It's not like I referred to someone obscure, like Lyndon Johnson. On the other hand, I recently got blank stares from the students when I referred casually to Alfred Hitchcock. With each passing semester I can just feel myself getting older!
With regard to the problem, GAZZA's opening comment was the solution I had in mind. I like that solution, because you can reason your way to within a whisker of it, leaving just a smidgen of trial and error. But the point is not to be legalistic, and I will accept any answer that is based on a reasonable interpretation of the problem statement. The more clever the better!
As someone who works at a university campus, I understand the feeling. But I hope the comment about Lyndon Johnson was snark. Popular culture is much more ephemeral than history. Expecting your students to remember the Bangles would be like expecting you or me to remember some group that had a couple of hit songs in the 1960s and was never heard from again. Since I am close to your age, of course I would be hard pressed for specific examples, because I don't remember those groups either. Even more recent groups sometimes escape my notice: I have seen a video consisting of excerpts of every song that was #1 on the US charts at some point in 1985 (which is the year I graduated from high school, so during the peak period of my pop music awareness), and there were songs and groups in that video of which I had no memory whatsoever.
There's a larger chance your students will have heard of Simon and Garfunkel, because S&G were much bigger and Simon remains active in pop music to this day. The comparable for us would be remembering people like Glenn Miller or Bing Crosby--they were big in the 1940s, and it's likely we have heard one or two of their big hits, but it's probably not music we actively seek out.
If you should ever find yourself playing the classic version of Trivial Pursuit against me, and you get to pick the category, choose Entertainment. Most of those questions, as I recall, deal with movies and movie stars from the 1950s and 1960s, about which I know very little. Once in a while I'll get lucky and draw a question about music--my knowledge there is a bit broader.
Jason Rosenhouse wrote (#14, September 7, 2016):
> [...] and I will accept any answer that is based on a reasonable interpretation of the problem statement [...]
Well, might the prescribed clock face be deformed (ever so silghtly)?
Then I'd suggest (guardedly only after the deadline)
to draw two intersecting straight lines resulting in four clock face sections with numbers (reading "left to right"):
6, 5, 4;
9, 0, 1, 2, 3; and
1, 1, 12, 1.