Great optical illusion
Here's a great optical illusion I found at this URL. I don't recall how I got to the page. It might have been from a twitter link.
This is a really cool illusion because, to my eyes, it looks like the circles cross. But, when I look closely I see that they don't.

What do you see?
MMM #17: The bishop and the king

Doug Hull was randomly chosen as the winner for MMM #16 over at Blinkdagger. Congratulations, Doug!
It's time for a new contest problem. As you may have noticed, I like counting and probability problems. I found an elegant one in an old book. I won't reveal what the book is until the end of the contest. Here's the problem:
Monday Math Madness #11
MMM #11 is a variation on MMM #9. I promise I won't do any more variations on this problem after this one!
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 11?
While you may use a computer program to verify your answer, show how to solve the problem without use of a computer.
MMM #9 was interested in divisibility by 8. This contest is interested in divisibility by 11.
I have a Rubik’s Revolution, courtesy of Techno Source, (or $10 Amazon.com gift certificate) to give to the winner. I’ll give more than one prize if I get lots of correct submissions.
I've changed rule #9 to encourage original solutions, which I'm much more likely to acknowledge:
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.
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.
3. 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.
4. The deadline to submit answers is Tuesday, July 29, 2008 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.)
5. The winner will be chosen randomly from all timely well-explained and correct submissions, using a random number generator.
6. The winner will be announced Friday, August 1, 2008.
7. The winner (or winners) will receive a Rubik’s Revolution or a $10 gift certificate to Amazon.com. For those of you who don't want a prize I'll donate $10 to your favorite charity.
8. Comments for this post should only be used to clarify the problem. Please do not discuss ANY potential solutions.
9. 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.
Monday Math Madness #9

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.
Monday Math Madness #7: We have a winner
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.
Monday Math Madness #7

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.
Janet’s solution to Monday Math Madness #5
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.
Warmup problems for Monday Math Madness #7
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.
Monday Math Madness #5

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.