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The results are in, and the winners for MMM #8 is. . . Heather Lewis! Thanks to everyone who participated! I wonder which MENSA book she'll choose?

The Answer

Here is Heather's explanation:

The answer is that Martin has a 29.6% chance of making it safely. As an aside, there's a 77.8% chance that at least one route is bully-free but, sadly, much of the time Martin doesn't pick the bully-free route.

I found two ways of solving this problem. One is brute force: There are 4 people and each picks a route (1, 2, or 3) at random, so there are 3^4=81 possible scenarios, all of which are equally likely. I listed those, and then for fun marked which ones had a possible open route (by just checking if the product of the three bullies' routes equaled 6). There were 63 of those. Then I marked in which cases Martin was on a route that none of the bullies took. There were 24 of those. So the probability of there being an open route at all is 63/81 (the 77.8%) and the probability of Martin being safe is 24/81 (the 29.6% chance).

Another way is to use probability. Martin will pick some route home, and the probability for each of Nelson and his brothers (Darryl and Darryl?) picking a different route than Martin is 2/3 [each]. This means there is a (2/3)^3 = 8/27 approx 0.296 chance that Martin gets home safely.

Special Thanks to this week's Sponsor, Mensa!

Before we introduce the problem for this week, we would like to take some time to thank our great sponsor, Mensa.

Mensa is the largest, oldest, and most famous high-IQ society in the world. The non-profit organization restricts its membership to people with high testable IQs. Members must score at the 98th percentile or higher of a standardized, supervised intelligence test.

The people at Mensa have provided Blinkdagger with a bunch of books that range from crossword puzzles, sudoku puzzles, and even a casino gambling guide! We will be handing out these books as the prizes for this contest. For the international participants, we are still offering the 10 dollar gift certificate to Amazon. See the Rules sections for more details! Without further ado . . . let the games begin!

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The results are in, and the winners for MMM #6 are . . . Henno Brandsma and Steven King! Hope you guys enjoy your gift certificate to The Art of Problem Solving! The turnout for this contest wasn't nearly as good as the previous one (48 submissions compared to 103), so that tells us that we need to do a bette job advertising and spreading the word. It also tells us that we need a better/more interesting math problem.

For the next contest, we will probably revert back to a single problem. Originally, I thought the idea of having an easier problem and a harder problem would be more appealing to people. Perhaps we'll try to do a double problem again next week! Any feedback would be appreciated!

Once again, we would like to thank our sponsors from The Art of Problem Solving! Stay tuned for the next Monday Math Madness!

Part I Explained

As explained by Efrit Freeq

Each step dilutes the fraction of mayonnaise to 500/510 of what it was at the
previous step. It starts at 100% mayonnaise and we need to dilute it to 50%.

So we need n such that (500/510)^n ~= 1/2

Hence n ~= ln(1/2)/ln(500/510) so n = 35 is as close as we can get.

Part 2 Explained

Explanation by Lieven Marchand

Let us denote by p(n) the number of ways Nortrom can eat n pounds of
beef. Since he will always start by eating either a 1 pound burger or
a two pound burger, p(n) obeys the following recursion:

p(n)=p(n-1)+p(n-2) with initial conditions p(1)=1 and p(2)=2.

These are the well known Fibonacci numbers and so the solution is

p(n)=1/sqrt(5) [((1+sqrt(5))/2)^(n+1) - ((1-sqrt(5))/2)^(n+1)]

This gives us p(17)=2584 and p(25)=121393.

Thanks to our Sponsor!

Welcome to Monday Math Madness! Before we introduce the problem for this week, we would like to take some time to thank our great sponsor, The Art of Problem Solving.

The Art of Problem Solving offers a challenging curriculum in problem solving mathematics for strong math students in grades 6-12. The bookstore also offers a variety of other books from prominent competitions, such as MOEMS, MATHCOUNTS, and the AMC. If you think the problems that you find here are fun, there are books at their website filled with countless problems that will keep your mind buzzing for years.

The people at The Art of Problem Solving have been kind enough to provide 25 dollar gift certificates to the winners of this contest. See the Rules sections for more details! Without further ado . . . let the games begin!

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Monday Mac Madness was the most successful contest we have conducted to date. In total, we received 103 submissions, which is twice as many as our original goal of 50. And since we received more than 50 submissions, we chose two winners for this contest!

Congratulations to Raymond Chong and Allan Wong for winning Monday Math Madness #4!

The Answer is 7!

The answer to the problem is 7. The two most common answers we received were 11 and 6. Some of the explanations were very clear and succinct while some of you struggled to explain your strategy.

Since the personnel at Blinkdagger are too lazy (or dumb) to write up a solution, they have instead chosen to explain the answer by choosing some of the best submissions. One of the clearest explanations came from Greg Zeigler, shown below:
"It would take 7 iterations to determine the three fastest macbooks. I'll include a visualization as well since this is tough for me to explain properly.

First, make five groups of five and run each group through 1 iteration. Discard the slowest two of each group since there are at least three machines faster than them.

Run an iteration using the fastest macbook from each group. Discard the two slowest, as well as the other two laptops remaining in each of their groups. Then go to the group that had the third fastest and discard the second and third laptops since they are slower than a macbook which is slower than at least two others.

For the same reason, you can discard the third laptop from the group whose laptop was second in the sixth iteration. You know that the first laptop from the sixth iteration (being the fastest of the fastest) is the fastest laptop you have, so you can set it aside. This leaves you with 5 laptops, enough for one more iteration.

The first laptop from the seventh iteration is the second fastest, the second is the third fastest, and as I already said the fastest laptop is the first laptop from the second iteration."

Another great explanation can be found here by Janet (the Geekmom).
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