## Winner for MMM #11

The prize for MMM #11 goes to Chao Xu. Here is Chao's solution.

A number divisible by 11 if it's alternating sum of it's digits is divisible by 11.

We only have 6 numbers, so we have a+b+c-d-e-f divisible by 11, where a to f are variables for each position. The largest possible number formed by a+b+c-d-e-f is 6+5+4-3-2-1 = 9, the smallest number is -9,

Then we have only one number that is divisible by 11 and between 9 and -9 is 0.

We have

a+b+c = d+e+f

a+b+c = (a+b+c+d+e+f)/2

and a+b+c+d+e+f = 21.

a+b+c = 10.5

It's not a whole number

It shows there is no number can be formed that's divisible by 11.

## 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 by11?

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

maypost 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: Winner!

We have a winner for MMM #9. It's Seb Perez-D. Congratulations! Drop me a line to arrange for your prize. And, check out Blinkdagger on Monday for their new problem.

A couple of people were confused about the deadline for submitting a solution. The deadline is 12:01 Tuesday Mountain Time, which is to say Tuesday morning. If you do a Google search for "time California" you'll know what time it currently is in California. If it's after 12:01AM the week after a contest is posted then you're late! Using an offset from GMT will get you into trouble since California changes its offset twice each year.