Math contest
Richard Berlin, who has been participating in Monday Math Madness! for quite a while, was picked as the winner for this contest by Random.org. Congratulations, Richard! I’ll be sending you your prize so send me your address. Stand by for another Monday Math Madness on Monday at Blinkdagger.
Blinkdagger and Wild About Math! are really stirring things up for Monday Math Madness (MMM) contests #13 and 14. And, Texas Instruments is giving away a VERY cool calculator (keep reading for more about the calculator), they’ll ship it internationally, plus we’ll allow anyone to win - even if you’ve won a prize in the MMM contest before. So, we expect submissions from everyone on the planet!
What’s different about the next two contests?
So, what are we doing differently? Well, we’re going to ask you guys and gals to submit your favorite MMM-caliber problems. That’s what you have to do be eligible to win MMM #13. We’ll pick the one we like best and then use it for MMM #14.
- If you submit the problem we pick for MMM #13 then you get one of the awesome calculators.
- If you solve the problem we announce in MMM #14 then you get an awesome calculator.
Last Friday Blinkdagger announced a winner for MMM #8. Here’s MMM #9:
Consider all of the 6-digit numbers that one can construct using each of the digits between 1 and 6 inclusively exactly one time each. 123456 is such a number as is 346125. 112345 is not such a number since 1 is repeated and 6 is not used.
How many of these 6-digit numbers are divisible by 8?
While you may use a computer program to verify your answer, show how to solve the problem without use of a computer.
We have a winner for this seventh contest. Congratulations, Brent Yorgey! I’m delighted that Brent, of Math Less Traveled, won this one because Brent gives so much to students and readers of his blog. Brent - Enjoy your $25 gift certificate from our kind sponsor for this contest, the Art of Problem Solving.
Click here to see Brent’s solution, two of them actually.
It’s time for Monday Math Madness #7. I love infinite series and I found today’s infinite series problem on the web. This is one of the most interesting of these kinds of problems I have run into. It’s challenging but not brutally difficult, so give it a try. I won’t reveal the source until the contest ends because the answer is posted with the problem.
Thanks to the sponsors for this contest, I have one $25 gift certificate left for the Art of Problem Solving. I also have a couple of Rubik’s Revolutions, courtesy of Techno Source. Depending on how many correct solutions I get I may give away two prizes.
As promised, here is Janet’s solution to Monday Math Madness #5.
Aside from 1 and 9, are there any perfect squares whose digits are all odd? Justify your answer.
No.
All perfect squares two digits and larger have at least one even digit.
Blinkdagger has announced the winners for contest #6. A little later today I’ll be posting contest #7.
In the meantime, here are a couple of warmup problems:
1. If a fish weighs one pound plus half its own weight, how much does the fish weigh? Do this problem quickly and without paper. I bet many of you won’t get it right the first time. It’s not a hard problem but it is tricky if you’re not paying attention. Try this problem out on your friends.
2. What is interesting about each of the following pairs of numbers: (2,2) and (5/2, 5/3)?
Stay tuned for Monday Math Madness #7, later today. It’s an interesting infinite series problem.
I received an email today from a young person asking for me to link to a Math Contest resource web-site. Being a lover of Math contests myself and recognizing the value of the site, I am delighted to not just provide a link but to write this blog post about it.
Below is the email I received. (Yes, I got permission from the author to publish it.)
Note: If you have a Math-worthy site to let the world know about, send me an email at wildaboutmath at gmail. If I like it I’ll post about it. Wild About Math! has a Google PageRank of 5, so if I write about your site, it’ll be very good for your PageRank.
We have a winner for MMM #5 and a new contest to be announced Monday at Blinkdagger.
We received 24 entries, one came after the deadline, so only the 23 that met the deadline were considered. All solutions were correct and well explained. Random.org picked the winner.
Congratulations, Janet! You win the $25 gift certificate to the Art of Problem Solving. They have outstanding quality Math books. I doubt you’d be disappointed with anything you buy from them.
Wow! We’re on our fifth contest already! Time flies.
For this contest we have a new prize: The Art of Problem Solving folks have donated several $25 gift certificates that can be used in their bookstore. These folks do an outstanding job of challenging and inspiring kids to learn Math. Check out their website if you haven’t already. I’ll give away either one or two certificates, depending on how many correct solutions we receive.
This problem may be challenging for some but I encourage you to view this problem as an exploration and to look for patterns that will help you to solve it. I picked this exploration-friendly problem in honor of the folks donating the prize; Art of Problem Solving is all about Math exploration.
