blinkdagger

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).

Best responses

1. 6 iterations.

   Only five machines can be tested simultaneously, so the 25 laptops must be randomly assigned to five groups of five.

   Then each group would be tested (iterations 1-5) and the fastest machine from each group would be set aside in a new group.

   This new group of five would have to contain the fastest macbook (it would rank first when rated against any four in the previous step).

   So the new group would be tested (iteration 6) and the fastest in the group would have to be it.

   After the AIME, this s**t is no problem. I’ve got some harder questions if you need them (you do).

2. Did this as a break from revising quantum…

   The answer I got was seven tests. To elaborate….

3. i think its 3 because…

   you need to do three to get the top three computers out of the twenty (by doing five each time)
   that’s it!

   but it seems too simple, could be completely off…and it probably is

4. The answer is clearly 6. After each test is conducted in iterations of 5, all the laptops are tested. The 6th iteration would simply rank the #1’s from 1 to 5, allowing the the ranks 1 to 3 to be established. But who cares they are macs. (GO UNIX)

5. Hope i get this week’s one right

   Thanks for the interesting puzzle.

   PS. If you offer a mac as a prize, yoru readership might potentially grow exponentially.

6. In the presumably low-probability event that I win,
   I’d like the Amazon voucher. I’d be prepared to accept
   a new Macbook Pro as an alternative.

Shout outs

These shout outs go to the people who submitted correct answers, but were not chosen to win the prize. If you included a link to your website/blog, I also hyperlinked it.

Raymond Chong
F Knauss
Il Park
Doug Hull
Kelvin Heng
Jordan Kwan
Chin Hui Han
Erin Rubin
Steve Maguire
Soren Furbo
Mickey Sheu
Janet (Geek Mom)
Steven King
Immanuel Mclaughlin
Lord Thomas Burbridge
Greg Zeigler
Partha Bhattacharya
Anneleen Van Geenhoven
Simon John Youssef Antoun
Yichao
SSDragon
Niall McPherson
Jing Yee Chee
Hostile Fork
Marijn Jongerden
Ingmar Dasseville
Hillel Gershuni
Dieter Shirley
David Sutoyo
Allan Wong
Jack Snoeyink
Gareth McCaughan
Lieven Marchand
Ada Yu
Craig Wiegert
Fred Shipley

Monday Math Madness #5

Don’t forget to check out Monday Math Madness #5 over at wildaboutmath.com this coming Monday!