CUNY Math Contest
March 12, 2009 am31 4:14 am
There’s a CUNY math contest… and now there’s a blog that follows the questions round by round: CUNY Math Challenge.
This is not an official site, but a blog following the action (as it were).
They’ve posted the five rounds. And once March 15 passes, they will start posting the solutions, and, I assume, some discussion on the problems.
It came to my notice as searchers jumped back and forth. Apparently, one problem has a solution in two parts, both of which have been puzzles here!
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4 Comments
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Hey there JD2718, I’m the Administrator over at the Challenge Blog. Thanks for the link. I’d be curious to know if you have any links or material related to when you featured the 2 problems you mentioned here. If you do, I’d love to link to it. If you’d post a comment here or on our blog that’d be great.
Thanks a lot.
Thanks for the pointer — it looks like they have some pretty good questions. I figure a talented high schooler should be able to work through most of those, but none of them is too easy to be worth tackling. I find the blog you linked difficult to read (grey on black? really?), so I didn’t try digging around it too much — are the later rounds supposed to get harder?
i like. i did the “world series” problem
in front of a live audience without a net
about ten years ago in a “finite math” class.
and sure enough, missed the “right” way
and did an awkward case-by-case
before seeing what “should have been obvious”.
good old counting problems.
the rolling ball is a vivid illustration of
“one-to-one correspondence does not imply isometry”…
one can imagine a student objecting that
since there’s always a point on the *inside*
diametrically opposite to the point on the *outside*,
the areas *must* be the same…
but then, when the *whole* outer ball is covered…
i haven’t even started on the square root…
We can simplify the solution given there for the world series problem: every sequence of games with 4 wins and 3 or fewer losses can be extended to a sequence of games with 4 wins and exactly 3 losses in a unique way. For example, WWLWW extends to WWLWWLL, WLLWWW extends to WLLWWW and WWLLLWW extends to itself. Thus, the answer is 2 * 7C4. The equality 3C3 + 4C3 + 5C3 + 6C3 = 7C4 is one instance of the so-called “hockey-stick identity.”