Blinkdagger hosting Monday Math Madness #4
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These guys had way too much fun writing up this problem. One of the things we do when we come up with these contest problems is that, when we don't make them up ourselves, we rewrite them so a Google search won't yield the answer. Quan and Daniel did an amazing job of rewriting this problem!
Anyway, head over to Blinkdagger and get going on the next problem. This one is more in their style - it's a logic problem.
We have a winner for Monday Math Madness #3
We have a winner for the third Monday Math Madness contest. It's Johan Potums. Congratulations, Johan! I'll be contacting you about your prize.
Blinkdagger has a very interesting new contest problem that they'll be posting Monday.
Seventeen people submitted entries. Everyone got the right answer and explained their answer well. Everyone realized (or figured out) that every year does indeed have a Friday the 16th. Proofs varied but everyone used modular arithmetic to demonstrate that regardless of what day of the week the year begins on, some month will have a Friday the 16th.
Monday Math Madness contest #3 ending soon
If you've not gotten your entry in for contest #3 you've only got through Monday night to do so. We've only gotten 10 correct submissions for this contest so far, so your chances of winning the prize are better than they were for the last contest.
It’s Monday Math Madness time! (contest #3)
We've completed two Monday Math Madness contests. Last Friday Blinkdagger announced that Joshua Zucker, director at Julia Robinson Mathematics Festival, was randomly selected as the winner of the 2nd contest. Now it's my turn to post a contest problem. Those of you who are astute readers may have noticed that I said the contest would be held the 1st and 3rd Mondays of the month and today is actually the 5th Monday of March. Well, there's enough enthusiasm about this contest so we'll just do it every other week. So, we'll do 26 contests per year rather than 24. We're nice that way!
Mathematician David Gale leaves legacy
Earlier this month UC Berkeley professor emeritus of mathematics David Gale passed away. Gale made a number of significant contributions to mathematics and he loved puzzles, games, and finding beauty in mathematics. Gale's daughter had this to say:
A very clever way to solve the first Monday Math Madness problem
On March 3rd Blinkdagger and I posted the first Monday Math Madness problem. On March 11th, after the first contest ended, I posted a couple of different solutions to the problem. Pat Ballew, even though he wasn't picked as the random winner, impressed me with a very clever solution to the problem that generalizes very nicely. He uses an approach called Markov state matrices, which I had never heard of. It seems to me that this approach is pretty similar to the one I posted from Richard Berlin. Pat and I exchanged several emails where he explained the method and here is my attempt to explain what Pat explained to me.
This was the problem:
A popular blog has just three categories: brilliant, insightful, and clever. Every blog post belongs to exactly one of the three categories and the category for each post is selected at random. What is the probability of reading at least one post from each category if a reader reads exactly five posts?
Pat's approach starts by creating a matrix that encodes the probabilities of going from one "state" to another as a new blog post is read. State just refers to whether 0, 1, 2, or 3 categories have been encountered after reading some number of blog posts. After one blog post has been read we are in state 1 (1 category has been read). After two posts have been read we may be in state 1 (if both blog posts are in the same category), or state 2 (if the two categories are different), but not in state 3 (you could not have encountered three categories after having read only two blog posts.)
Monday Math Madness: We have a winner!
For the very first Monday Math Madness contest we got 13 submissions. Of the 13, 6 were correct. For the record, I solved the problem by enumerating the various cases where 3 categories were represented and computing and adding their probabilities. I also verified my solution to the problem by writing a computer program to enumerate all 243 (3^5) permutations of 3 categories and 5 blog posts and count the ones were all 3 blog categories were represented. So, I'm pretty confident I got the right answer 
Keep those Monday Math Madness answers coming!
Four of you have submitted answers to Monday Math Madness so far but only two of them are correct. Remember, the Blinkdagger guys and I are picking a good random answer, not the first good answer. So, you've all still got a chance to win $10 to Amazon.
Heather, from the 360 blog, gave the problem to one of her classes to solve, and told them how to submit their answers. So, if you've got Math students who have learned some probability then have them try their hand at this problem. It's not super difficult but it's tricky enough that only half the solutions submitted were correct.
You all have until Sunday night to get your submissions in. And, if you don't like this problem or miss the deadline, Blinkdagger will be running the next contest, starting in 12 days.
Happy problem solving!
Monday Math Madness is here!
I was contacted last week by a couple of Matlab geeks, Quan and Daniel, who have started a 
blog, Blinkdagger, about co-sponsoring an ongoing Math contest. Now, I was a little confused because I think these guys are engineers, and I didn't know that engineers took Math classes (Maybe these guys learned Calculus in the 4th grade.) Anyway, their email got past my spam filter so I figured this relationship was meant to be! So, (drum roll, please), Blinkdagger and Wild About Math! have teamed up to post fun Math puzzles on the 1st and 3rd Monday of every month. There'll be a prize for the best combination of randomly selected right answer plus good explanation of how you got your answer.
What do these words have in common?
Ok, so this is not Math-related but I did discover it in the book Fun With Mathematics. The book, by the way, which is out of print and only available used, is definitely worth a couple of bucks plus the shipping cost.
So, what do the following words have in common?
playgrounds
dumbwaiters
workmanship
republicans
sympathizer
Yes, they each have 11 letters but there's something else interesting about each of these words. Can you see what that is? Are there other words you can add to this list?
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