Wild About Math! Making Math fun and accessible


MMM #25: Monotonous Monday Math Madness

[ See below for MMM #25 ]

Blinkdagger has announced the winner for MMM #24 and I've got a new problem so keep reading.

Rich Berlin noticed the relationship to MMM #1 and submitted an interesting solution, based on Pat Ballew's Markov chain process which I wrote about in March.

The technique for solving this is the same as for MMM #1, except the numbers are a bit nastier to work with.

I solved MMM #1 using a decision tree, but as a result of reading Sol's blog I learned the Markov state approach and am happy to use that instead.


Monday Math Madness #9

Monday Math Madness

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.


A very clever way to solve the first Monday Math Madness problem

On March 3rd Blinkdagger and I posted the first Monday Math Madness problem. On March 11th, after the first contest ended, I posted a couple of different solutions to the problem. Pat Ballew, even though he wasn't picked as the random winner, impressed me with a very clever solution to the problem that generalizes very nicely. He uses an approach called Markov state matrices, which I had never heard of. It seems to me that this approach is pretty similar to the one I posted from Richard Berlin. Pat and I exchanged several emails where he explained the method and here is my attempt to explain what Pat explained to me.

This was the problem:

A popular blog has just three categories: brilliant, insightful, and clever. Every blog post belongs to exactly one of the three categories and the category for each post is selected at random. What is the probability of reading at least one post from each category if a reader reads exactly five posts?

Pat's approach starts by creating a matrix that encodes the probabilities of going from one "state" to another as a new blog post is read. State just refers to whether 0, 1, 2, or 3 categories have been encountered after reading some number of blog posts. After one blog post has been read we are in state 1 (1 category has been read). After two posts have been read we may be in state 1 (if both blog posts are in the same category), or state 2 (if the two categories are different), but not in state 3 (you could not have encountered three categories after having read only two blog posts.)


Monday Math Madness: We have a winner!

For the very first Monday Math Madness contest we got 13 submissions. Of the 13, 6 were correct. For the record, I solved the problem by enumerating the various cases where 3 categories were represented and computing and adding their probabilities. I also verified my solution to the problem by writing a computer program to enumerate all 243 (3^5) permutations of 3 categories and 5 blog posts and count the ones were all 3 blog categories were represented. So, I'm pretty confident I got the right answer :)


Probability and divisibility by 30

Well, probability is on my mind - this sounds like a great song title, no?

I've got another probability problem, this one presented to me by my brother Abe as I called him from the airport half way to my awesome vacation in Hawaii. I chewed on it for a little while then called him back with the answer, got his answering machine and left the answer there to impress him. He didn't call me back but I think I got the answer right. Maybe that's why he didn't call?

Here's the problem:

What is the probability that a 10 digit number will be divisible by 30?

The problem was really stated as asking how many ten digit numbers are that are divisible by 30, which is in essence the same problem.

This is a neat problem in that it involves several aspects of solving Math problems and has a number of dimensions to it that lead to a pleasant exploration. Some ideas on approaching this problem:

  1. Once must assume that a 10 digit number has no leading zeros, i.e. that the first digit is between 1 and 9. How does this special case for the first digit affect the probability? Without this consideration someone might hastily and incorrectly conclude that the probability is real close to 999,999,999 / 30.
  2. How many 10 digit numbers are there? How would you count them?
  3. If you take the number of 10 digit numbers and divide that number by 30 do you then get the right answer? Why or why not?
  4. What is interesting about the number 30? How would you test for divisibility by 30? Hint: Factor 30 into primes.
  5. Does the fact that 30 ends in zero lead to a simplification of the problem?
  6. Can you solve this problem for numbers with fewer digits? How many single digit numbers are there divisible by 30? How many 2 digit numbers? How many such 3, 4, and 5 digit numbers are there?
  7. What if the problem asked about divisibility by some other number, maybe one that fewer or more factors? Might it then be easier to utilize an approach that didn't take into consideration the factors of the number we're testing divisibility for?
Filed under: Probability 1 Comment

Fun calendar probability problem

I picked up a little gem from Math Wonders to Inspire Teachers and Students, byMath Wonders to Inspire Teachers and Students Alfred Posamentier. The book, by the way, is outstanding. Posamentier has become one of my heroes for writing books that awe and inspire students and teachers to get, well, Wild About Math! Someday I will review this, and other, outstanding books that make Math accessible and fun.

Here's the little problem:

What's the probability that 4/4, 6/6, 8/8, 10/10, and 12/12 all fall on the same day of the week?

It's fun to see what people come up with, especially if they give you an answer off the top of their heads.

A fun little exploration is why the answer is what it is. This exploration is a nice first step in doing calendar arithmetic.

Filed under: Probability No Comments