Friday, February 20, 2009

Updates and A Challenge Problem From Our First Contest




Problem 5 (2 pt question)




Totally_clueless said...

Alternative solution to the Math problem of the day (2/20/09):

To prove: P= prod_(i=1)^(k) (m+i) is divisible by k! for all m > 0.

As mathematicians like to do, I will just pull something out of thin air to prove the assertion.

P/k! = (P.m!)/(k!m!) = (m+k)!/(k!m!) = C(m+k,k) = n, where n is an integer. QED.


Dave Marain said...

Nice, elegant, and much more satisfying than the indicated inductive proof which appeared to have some flaws, not to mention notational issues!

Another approach based on fundamental divisibility concepts (remainders, etc), which should be part of a middle schooler's repertoire, is the following idea:
(i) Given any set of 2 consecutive integers one is divisible by 2;
(ii) In any set of 3 consecutive integers exactly one is div by 4 (again consider remainders!);
(iii) In any set of 4 consecutive integers, exactly one is div by 4, etc., from which the factorial statement follows easily.

Now the general proof of the assertion I'm making does require induction or an argument similar to yours, but the idea behind this argument may be more accessible to 7th-10th graders.

Gee, if these Problems of the Day generate this much discussion, I'll be able to take a break from writing my own investigations for a while!

So how do you feel about the contest problem I posed? Straightforward, mechanical, prosaic?? There is a catch in there if one uses a particular algebraic set-up so I'm interested in reactions. I will post the answer to the 2 parts shortly...

Dave Marain said...

Silly typo in my divisibility argument which I'm sure everyone noticed!
In any set of 3 consec. integers, exactly one is div by 3...

Totally Clueless said...

Hi Dave,

Well, you could say " In a set of 3 consecutive integers, at most one is divisible by 4"

Also, I like the way the contest problem was structured. The students have to use the information provided in the preamble to calculate c, and then move on to the rest of the problem. The two parts have their own nuances, and the solutions are gratifying (non-messy!!).

Had you posted a very similar problem on the blog earlier?


Dave Marain said...

I've had several similar quadratic function problems on the blog although none quite like this one. I wrote this from scratch with the intents that you observed. I re-worked the problem many times to include those little twists.

My underlying purpose in writing these problems is to provide learning experiences for students. I believe good problems can 'teach' as well as or even better than all the explanations we provide. I should mention that 2-3 teams obtained the correct answer to both parts, although I didn't ask for an explanation so I'm really not sure if they recognized all of the subtleties. A couple of teams didn't pick up all of the clues, particularly that P and Q have to be distinct points so that Q cannot be (0,3).