Quan and Daniel have posted the answer to MMM #28 at Blinkdagger so now it's time for MMM #29.
I call this one the "mother of all clock angle problems." Some of you may have run into those problems where you have to figure out the angle between the two hands of a clock at weird times. Well, I took that problem as a starting point and added a twist to it.
Consider a 12-hour analog clock with two hands and a round face. Consider the angle between the two hands at any given time and, when the angle between the hands is not 180 degrees, take the smaller of the two angles. Thus, at 12:00 the angle between the two hands is 0 degrees. At 3:00 and at 9:00 it's 90 degrees.
If we me measure the angle between the two hands at each of the 61 consecutive minutes between 12:00 and 1:00 inclusively, what is the sum of those 61 angles?
Remember to show your work.
Here's an interesting equation:
[pmath size=12] 50 1/2 + 49 38/76 = 100 [/pmath]
It's interesting because the left side of the equation uses each of the digits between 0 and 9 exactly once.
Here's another example:
[pmath size=12] 97 30/45 + 2 6/18 = 100 [/pmath]
Can you find other such equations where the left side has whole numbers and fractions which sum to 100?
For those of you so inclined, can you get a computer to find all solutions?
Texas Instruments (TI) and Wild About Math! are collaborating on a new exciting contest. It’s about Math but in a different, but still creative way. TI will be donating two of their high-end CAS calculators to two of you. This is a truly advanced calculator. The TI-Nspire CAS state-of-the-art computer algebra system is useful for step-by-step, arithmetic, algebraic and calculus calculations and helps calculator users see the patterns and grasp the math behind formulas.
You have two weeks to get your entries in.
What’s the contest about? See the original post.
You have until Monday night at midnight to submit your answer to MMM #28 at Blinkdagger.
Here's the problem:
The Problem Statement
Daniel: Hey Quan, we just received a shipment of the latest Green MacBook Pros from Apple Inc. Unfortunately, they heard about what we did to the last shipment, so they only sent 2 this time. One for you, and one for me.
Quan: Hm, what are you going to do with your laptop?
Daniel: Well, I just moved into a new apartment complex that is 36 stories high. I’ve always pondered how durable these Green MacBooks really are. And I was wondering what the highest floor is from which I can drop this MacBook, and still have it operational . . .
Quan: Well, since you only have one MacBook, you could start at the first floor. If you drop it there, and it doesn’t break, you can move on up to the second floor. Using this method, you can work your way up to the 36th floor. The worst case scenario is that it could take 36 iterations, but eventually you’ll find out the highest floor that the MacBook can survive from!
Daniel: Yes, I could do that, but my apartment complex doesn’t have an elevator, and I would get tired of running up and down the stairs all day. Now, if I had two Green MacBooks at my disposal, I could perform this task in a much more efficient manner.
Daniel: Don’t worry Quan, I have it all figured out. In the worst case scenario, it will only take us ____ iterations to figure out what the highest floor is!
How many iterations do Quan and Daniel have to perform to determine the highest floor that a Green MacBook can be dropped and still be operational?
You may assume the following:
- A MacBook that survives a fall can be used again.
- A broken MacBook must be discarded.
- The effect of a fall is the same for all MacBooks.
- If an MacBook breaks when dropped, then it would break if dropped from a higher window.
- If an MacBook survives a fall, then it would survive a shorter fall.
- It is not ruled out that the first-floor windows break MacBooks, nor is it ruled out that the 36th-floor windows do not cause a MacBook to break.
Random.org has picked Dooglius as the winner of MMM #27. Congratulations, Dooglius!
The answer to the problem is 1/(w^3-w^2-w+1), or 1/((1-w^2)(1-w))
Tanaeem M Moosa had a very clever demonstration. Check this out:
s=1 + (w+w^2) + (w^2+w^3+w^4) + (w^3 + w^4 + w^5+ w^6) + (w^4 + w^5+ w^6 + w^7 + w^8) + ...
w*s= (w) + (w^2+w^3)+ (w^3+w^4+w^5)+ (w^4+w^5+w^6+w^7)+...
or, s(1-w) = 1/(1-w^2)
Blinkdagger will have their next contest on Monday so check it out there!
William Wallace sent me the following email so I'm inspired to promote his puzzle:
I came across your math blog, and thought you might be interested in a puzzle I came up with.
To see how it works, go to (http://blog.coincidencetheories.com/?p=1522). You probably can figure it out first, then go verify. You can blog on it. I released it to the public domain so you don't need to give credit, but you may. Leave a pingback--I honor them. The above website has a pdf version.
How to play:
In order to calculate the number your child picked, you must be able to add two digit numbers in your head. If you can do this, think about the final sum, and start off by setting it to 0 in your mind. As you show each page, if your child picks a blue(cross) or green(square), find the lowest value on the page that is blue(cross), or green(square), whichever she picked, and add that to your sum. If your child picked red(circle), simply add zero (0) to your sum. Repeat this for each of the next three pages. After the forth page, the sum in your mind is the number that your child picked. Announce it. Your child will think you're a mind reader.
For a long time I've been very interested in Mathematica as a tool for mathematical exploration. But, I've never been interested in paying nearly $2,500 for the retail version. (Yes, I'm aware that there are steep educational discounts but I'm not a students so those don't apply to me.) Recently, Wolfram Research announced the Mathematica 7 Home Edition for $295. Wow! I'm now very excited about the value of Mathematica, especially in playing with Math ideas. And, if you were wondering, at only 12% of the Professional Edition, Mathematica 7 Home Edition is NOT a stripped down or time-restricted version of the software. It's the full-blown Mathematica 7 with the only restriction being that Mathematica 7 Home Edition cannot be used for teaching, research, or work.
I received a review copy of the Home Edition from Wolfram and I will be reporting on what I discover about the program.
If it's not immediately obvious to you that today is the square root of Christmas, then check this out:
Texas Instruments (TI) and Wild About Math! are collaborating on a new exciting contest. It's about Math but in a different, but still creative way. TI will be donating two of their high-end CAS calculators to two of you. This is a truly advanced calculator. The TI-Nspire CAS state-of-the-art computer algebra system is useful for step-by-step, arithmetic, algebraic and calculus calculations and helps calculator users see the patterns and grasp the math behind formulas.
What's the contest about? Read on.
MMM #26 was a tough one. The answer: 3 planes suffice! Read more here.
Here's the next Monday Math Madness! problem. I decided to do an algebra one for a change.
Given that the absolute value of w < 1. Simplify this expression: 1 + (w+w^2) + (w^2+w^3+w^4) + (w^3 + w^4 + w^5+ w^6) + (w^4 + w^5+ w^6 + w^7 + w^8) + ...
Here are the rules for the contest:
1. Email your answers with solutions to mondaymathmadness at gmail dot com.
2. Only one entry per person.
3. Each person may only win one prize per 12 month period. But, do submit your solutions even if you are not eligible.
4. Your answer must be explained. You must show your work! Wild About Math! and Blinkdagger will be the final judges on whether an answer was properly explained or not.
5. The deadline to submit answers is Tuesday, March 10, 12:01AM, Pacific Time. (That’s Tuesday morning, not Tuesday night.) Do a Google search for “time California” to know what the current Pacific Time is.)
6. The winner will be chosen randomly from all timely well-explained and correct submissions, using a random number generator.
7. The winner will be announced Friday, March 13, 2009.
8. The winner (or winners) will receive a Rubik’s Revolution or a $10 gift certificate to Amazon.com or $10 USD via PayPal. For those of you who don’t want a prize I’ll donate $10 to your favorite charity.
9. Comments for this post should only be used to clarify the problem. Please do not discuss ANY potential solutions.
10. I may post names and website/blog links for people submitting timely correct well-explained solutions. I’m more likely to post your name if your solution is unique.