Penny is for Penalize

The elementary school where my boys attend is hosting a fund raising contest where students bring their change to donate. Instead of accumulating donations to win for their own grade, they are putting their change in other grades canisters. The grade with the lowest weight in donated change at the end of the fund-raising wins. Pretty innovative, I think. The kids seem motivated, so fortunately for the school they are not colluding to all bring nothing which would keep their weights down. Instead they are piling in the change especially those with siblings in other grades.

Strolling through the halls, I overheard a conversation where teachers were wondering if anyone had found out which coins weighs the most so they could give more of those.  Of course, that got me thinking.  The obvious answer would be that the larger coins like the half-dollar or presidential dollar would be the heaviest.

Although, don’t you really want to have the most weight for your money?  Sure, I could put in 10 presidential dollar coins, but which would weigh more, 10 dollars in half dollars, 10 dollars in quarters, or 10 dollars in dimes?

So, of course, I had to know the answer.  Checking out the U.S. Mint, I learned the following weights for each of the coins:

1 penny = 2.5 g  which means $1 in pennies is 250 g.
1 nickel = 5 g which means $1 in nickels is 100 g.
1 dime = 2.268 g which means $1 in dimes is 22.68 g.
1 quarter = 5.670 g which means $1 in quarters is 22.68 g.
1 half-dollar = 11.340 g which means $1 in half-dollars is 22.68 g.
1 dollar (coin) = 8.1 g which means $1 in a single coin is 8.1 g.

In spite of the larger weights for the larger coins, you are still much better off dumping in those pennies.  I did learn a pretty interesting fact though: A dollar in dimes weighs the same as a dollar in quarters which also weighs the same as a dollar in half-dollars.  Pretty cool!

Second Chance Exam

imageHave you ever heard of a “second chance exam”?  I came across the concept for the first time in an article at Faculty Focus, Revisiting Extra Credit Policies.

Here’s how the author explains it:

The instructor attaches a blank piece of paper to the back of every exam. Students may write on that sheet any exam questions they couldn’t answer or weren’t sure they answered correctly. Students then take this piece of paper with them and look up the correct answers. They can use any resource at their disposal short of asking the instructor. At the start of the next class session, they turn in their set of corrected answers which the instructor re-attaches to their original exam. Both sets of answers are graded. If students missed the question on the exam but answered it correctly on the attached sheet, half the credit lost for the wrong answer is recovered.

I currently have a standing policy in all of my classes that allow students to correct missed problems on a test after it has been graded.  They’ll receive a bonus point on their exam grade for every correctly revised problem.  Instead of a flat bonus, this gives the most reward to students who put in the most work in the corrections. 

I used to do a flat 10 point curve for corrections.  At one point I was having students hand a test notebook at the end of the term.  The notebook contained corrected versions of their tests and they were rewarded with 3 bonus points on the their final average.

I’m considering trying this new approach, the “second chance exam” because it requires students to assess what they know, put in the work of correcting a problem and it also reduces the amount of time it takes to get a final grade into the grade book.  Right now, students take a test, then get it back the next class, then turn in corrections after that, and then I eventually return their corrections.  This new way, I collect the second chance exam the next class after the exam and then return the fully graded exam after that.

Of course, a sizeable percentage of fellow faculty would probably argue that extra credit only encourages laziness and procrastination on the part of the students but if the opportunities can be manipulated into a learning experience, isn’t that better than not learning at all?

CAMT 2011 – Our Nerd Vacation

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My wife, Lori, and I have come to Grapevine, TX to attend one of the largest state-level mathematics conferences in the U.S.  We are at the Conference for the Advancement of Mathematics Teaching (CAMT) which is a joint conference held by the NCTM, MAA and TASM.  (Google them if you really have to know what those acronyms stand for)

The conference is primarily for K-12 math teachers but there are few like me here that participate or supervise in the area of teacher education.  I’m here to learn two things, better techniques to teach our up and coming teachers and changes coming due to STAAR and the EOCs.  (The tests replacing the current TAKS.)

The venue is the Gaylord Texan Resort and Convention Center in Grapevine, Texas.

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We have a beautiful room with an excellent view of the atrium.

IMG_1180 Stitch

 

Now this is more for me as a reminder than for the readers but here are the talks I have attended so far:

 

Day 1 (Monday, July 18)

First Timer’s Session (8:00 – 9:00)

Algebra I Activities With The TI-NspireTM Handheld – Andi Parr, Region 13 ESC (9:15 – 10:15)

The Role of Inquiry Teaching Methods in Secondary Mathematics Classrooms – Mark Daniels, University of Texas at Austin (10:30 – 11:30)

Ignite Session! – Tim Pope, Key Curriculum Press (11:45 – 12:45)
Note: This session had a unique format.  Nine speakers each had 5 minutes to present.  They each had 20 powerpoint slides that would advance ever 15 seconds automatically.  This year’s speakrs included; Pam Harris, Paula Moeller, Michelle King, James Epperson, Amber Branch, Emma Trevino, Cindy Schneider and Cindy Schimek.

Exploring AP Caluclus Activities with the TI-NspireTM – Noe Medrano (3:30 – 4:30)

 

Day 2 (Tuesday, July 19)

Small Group Instruction in the Secondary Classroom – Richard Yoes, Joda Mendoza, Pasadena ISD (8:00 – 9:30)

The New TI-Nspire Navigator SystemTM – Holly Larson, McKinney ISD (9:15 – 10:15)

Big Gains from Small Struggles – Cathy Seely, Charles A. Dana Center (10:30 – 11:30)

Math Curriculum Makeover – Dan Meyer, Author (11:30 – 1:00)
KEYNOTE SPEAKER

 

…more to come later

Book Review: Evolving in Monkey Town

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Evolving in Monkey Town: How a Girl Who Knew All the Answers Learned to Ask the Questions

by Rachel Held Evans.

My Rating:      didn't like itit was okliked itreally liked it (my current rating)it was amazing

Status:    Read from June 25 to 30, 2011

My Review:

Based on a popular blog, this book walks through a twenty-somethings evolution of faith from a fairly fundamentalist background to a more liberal and reformed approached to Christianity. I was surprised a just how much of her experiences mirrored my own. She definitely poses the more difficult questions that a Christian must deal with today. I hoped to have her tackle the questions more directly than to just have her cherish the ability to pose them. Nevertheless, it was a good read and I would recommend it to any young adult who has struggled with doubt.

Arcs in a Square (or Snakes on Plane)

image In a recent MAA publication, Shai Simonson, attempts to bring the joys and excitement of the world of mathematics to the non-technical reader.  In Rediscovering Mathematics: You Do the Math, Simonson covers a wide array of topics ranging from number theory to the application of probability in sports, casinos and gambling.  I have added the book to my reading list and you might want to take a look at the article that begins his text, “How to Read Mathematics.”

One of the problems from the book was posted over at Math Mama Writes… and when a puzzle like this piques my interest, I’m at its mercy until I figure it out.  Thanks to a recent illustration I made in calculus last week and a cartoon that reinforced my perspective, I have a new motto for next year’s courses:

Math problems aren’t solved, they are conquered!

Well, this problem below was one that I had to defeat.  I went to battle with it and after losing a few skirmishes (i.e., trying approaches that failed) I finally beat it into submission.  From now on, when I see that feisty integral that won’t behave or a simple number puzzle whose pattern defies identification, I’ll strap on my armor (or sweater vest), grab my sword (or calculator) and wage full-out war on that problem.  No problem is safe!

Arcs in a Square (or Snakes on a Plane)

Given the square ABCD, with side length 4 and circular arcs centered at each vertex, find the area of the region at the center – without using calculus.

And by the way, the Snakes on a Plane reference is my own and I’ve not ever heard this problem referred to in this way but it sounded good to me (arcs [tex]\approx[/tex] snakes, square [tex]\approx[/tex] plane).  However, I am a strange duck.

I’ll post a couple of different approaches that conquered this problem in later posts.

The Locker Problems Solved

imageA few days ago, I posted a couple of problems that I had stumbled across over at Math Jokes for Mathy Folks.  If you haven’t already, take a moment to read the puzzles:  The Locker Problems.

Problem 1

To start with, we can simply begin walking through process by hand.  Locker 1 will begin closed, then the 1st person will come in an open it.  After that, no one touches the 1st locker so we know it stays open.

Locker 2 is opened by the 1st person, closed by the 2nd and it stays closed.  Locker 3 is opened by the 1st person, left alone by the 2nd and then closed by the 3rd.  This is going to get tedious if I keep explaining in words.  Lets use a table.  If a locker is open, I’ll put a “o”, if it’s closed I’ll put a “c”.

 

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This is also going to get tedious if I want to do this for 1000 lockers and 1000 students.  We definitely need to find a pattern. 

My next step was put together a lazy little Matlab script to do the same exact thing as my table. (See here.)  Using the “spy” function in MATLAB allows me to see the sparsity pattern of the matrix.  The open lockers are represented by 1’s and the closed ones by 0’s.  Here’s what it looks like for 100 lockers and 100 students.

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There is definitely a pattern!  Notice those open lockers form the bands you see in the picture.  So why do some end up open and some end up closed?

The easiest way to see why is to consider what happens to a single locker.  For example, think about locker 24.  When does it change state?  Obviously, person 1 opens the locker, person 2 closes it, person 3 opens it, person 4 closes, person 5 does nothing, etc.  Notice that if the locker number, 24, is divisible by the person number, then the state changes:

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The number 24 has 8 factors, that is, 8 numbers that divide evenly into it.  So the key is that any locker number with an even number of factors will end up with closed and any with an odd number will end up open. So what numbers have an odd number of factors?

At first thought, it seems that all numbers should have an even number of divisors since they always occur in pairs: 

24 = 1 x 24
24 = 2 x 12
24 = 3 x 8
24 = 4 x 6

Consider a number like 36, though.

36 = 1 x 36
36 = 2 x 18
36 = 3 x 12
36 = 4 x 9
36 = 6 x 6 <—- Ah ha!!

Perfect squares are the only numbers that can have an odd number of divisors.  So locker numbers which are perfect squares (i.e., 1, 4, 9, 16, 25, …) will remain open while the rest stay closed!! That matches exactly with the table above and the graph from MATLAB.

All we need to know is how many perfect squares are less than 1000. 

[tex]\sqrt{1000} \approx 31.6222[/tex]

ANSWER:  There will be 31 open lockers

Problem 2

In the second version of the problem, we have only 30 students and some are absent.  In fact, we need to determine which ones were absent so that all the lockers except the first on remain closed.

Starting in a similar fashion as before, we could build a table by hand and determine who would have to be sick.  For example, the 1st student is clearly present since the first locker remains open.  After that, the 2nd and 3rd are both present since those lockers end up closed.  The 4th person must be absent since if they were present then the 4th locker would be open. 

So any locker number that has an odd number of factors (only counting factors if that person is present) must not be there.  For example, consider the 8th locker.  The number 8 has 4 factors but one of them is 4 so it actually only has 3 factors when counting those present.  So 8 must be out sick. 

Following this pattern, we eliminate all multiples of 4.  Then when we get to the 9th locker, they will also absent.  So will all multiples of 9.  Continue this pattern until you reach 30.

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The numbers that are not shaded are called squarefree since they have no repeated factors when you they are prime factored.  (Read more about squarefree numbers.) The absent folk are the "non-squarefree” numbers, also known as squareful.

ANSWER: There are 11 squareful numbers less than 30 so 11 students are absent.

That was fun!

The Locker Problems

image Problem 1

Every day, 1000 students enter a school that has 1000 lockers. All of the lockers are closed when they arrive. Student 1 opens every locker. Student 2 closes every other locker. Student 3 then “changes the state” of every third locker – that is, he opens it if it’s closed, and he closes it if it’s open. Student 4 then changes the state of every fourth locker, Student 5 changes the state of every fifth locker, and so on, so that Student n changes the state of everynth locker.

Which lockers are open after all 1000 students have finished opening and closing lockers?

 

Problem 2

Every day, 30 students enter a room with 30 lockers. All of the lockers are closed when they arrive. Student 1 opens every locker. Student 2 closes every locker. Student 3 then “changes the state” of every third locker — that is, he opens it if it’s closed, and he closes it if it’s open. Student 4 then changes the state of every fourth locker, Student 5 changes the state of every fifth locker, and so on, so that Student n changes the state of every nth locker.

One day, some students are out sick. Regardless, those present repeat the process and just skip the students who are absent — for instance, if Student 3 were absent, then no one would change the state of every third locker.

When they finish, only Locker #1 is open, and the other 29 lockers are all closed. How many students were absent?

 

HT: Math Jokes for Mathy People

Solutions have been posted:  The Locker Problems Solved