MathNotations

Posted by Dave Marain at 11:11 AM 0 comments

Labels: algebra, Common Core, explorations, investigations, iteration, precalculus, recursion

Posted by Dave Marain at 5:58 PM 0 comments

Labels: algebra, Common Core, investigations, radical equations, SAT-type problems, standardized assessment

Posted by Dave Marain at 6:43 AM 0 comments

Labels: algebra, SAT-type problems

Posted by Dave Marain at 8:22 AM 0 comments

Labels: advanced algebra, algebra, connections

Posted by Dave Marain at 6:13 PM 0 comments

Labels: algebra, SAT-type problems

CONTEST IS OFFICIALLY OVER AND THE WINNER IS ----- NO ONE! Guess I should have offered a 64GB 3G IPad! to be awarded on Black Friday...


The floor is now open for David, Curmudgeon, and my other faithful readers to offer their own solutions.


And the next contest is...




This is a contest so students must work alone and this needs to be verified by a teacher or parent. No answer will be posted at this time. Deadline is Wed 11-17-10 at 4 PM EST.






Here's a variation on the classic motion-type problems we don't see as often in Algebra I/II but still appear on the SATs. I found this in some long-forgotten source of excellent word problems to challenge NINTH graders!

Barry walks barefoot in the snow to school in the AM and back over the same route in the PM. The trip to school first goes uphill for a distance, then on level ground for a distance and finally a distance downhill. Barry's rate on any uphill slope is 2 mi/hr, any downhill slope is 6 mi/hr and 3 mi/hr on level ground. If the round trip took 6 hours (hey, these are the old days in the 'outback'), what was the total number of miles walked?


First five correct answers with complete detailed solutions emailed to me at dmarain@gmail.com will receive a downloaded copy of my new book of Challenge Problems for the SATs and Beyond when it becomes available. Both the student and teacher(s) will receive this. (Illegal to reproduce or send electronically!). Read further...

Submission by email must include (Number these in your email and copy the validation as well).


1. Answer and complete detailed solution. If answer is correct but method is sketchy or flawed, the submission will be rejected.
2. Full name of student
3. Grade of student
4. Math course(s) currently taking
5. Math teacher's name(s) and parent's name(s)
6. Name, Complete Address of School; Principal's Name & Email address (if known)
7. Email addresses of teacher(s), parents, student
8. Phone number (in case I need to call you) - Optional
9. How your or your teacher or parent became aware of MathNotations.




VALIDATION


I certify that my student (child) did the work independently.




--------------------------------------------------------------------------------


Name of Teacher or Parent (if work done at home)



"All Truth passes through Three Stages: First, it is Ridiculed... Second, it is Violently Opposed... Third, it is Accepted as being Self-Evident." - Arthur Schopenhauer (1778-1860)

"You've got to be taught To hate and fear, You've got to be taught From year to year, It's got to be drummed In your dear little ear. You've got to be carefully taught." --from South Pacific

Posted by Dave Marain at 12:41 PM 5 comments

Labels: algebra, contest, math contest, online math contest, rate-time-distance problems

Ok, so I don't have another anecdote from my grandson today so I'll have to get back to mathematics -- problem-solving, teaching and learning.

The video below deals with an algebraic equation in 2 variables which should be straightforward for your stronger Algebra II or Precalculus student. But will it be? I invite you to predict how many in your classes will answer it correctly, then try it out. After all, it is multiple choice, so statistically some should get it right by some means or other!

Hopefully, the purpose of the problem and the video will become clear to all of you. If we want our students to demonstrate better reasoning and an understanding of important ideas in math, we need to feel comfortable in teaching for meaning and understanding. This doesn't mean we stop teaching algorithms and procedures, however. Exactly what all this means and how to do it is the reason for this blog. I certainly never claimed to know the answers or any other mystical secrets. I only know that I never gave up trying. Sometimes my efforts failed miserably, but I hopefully learned from these attempts.

It would mean a lot to me if you share your thoughts here or on my You Tube channel, MathNotationsVids, where you will find my other videos.

[埋込みオブジェクト:http://www.youtube.com/v/xGKmPHfsUaU&hl=en_US&fs=1&]


Note: Another subtle point I should have made in the video---
y(x-4) = 0 → y = 0 OR x = 4
It is important for us to stress this point and distinguish it from "AND" logic. If the equation were in the form: y² + (x-4)² = 0, we would have (y = 0) AND (x = 4), whose graph would be the single point (4,0). Another instance where an exercise on the board can lead to a rich, fruiful and profound discussion. If all of this is seen as taking too much time away from content, remember this is precisely the kind of change in curriculum and instruction that Prof. Schmidt has been trying to tell us about for over 15 years! Well, I'm preaching to the converted, aren't I...
-------------------------------------------------------------------

"All Truth passes through Three Stages: First, it is Ridiculed...
Second, it is Violently Opposed...
Third, it is Accepted as being Self-Evident."
- Arthur Schopenhauer (1778-1860)

You've got to be taught
To hate and fear,
You've got to be taught
From year to year,
It's got to be drummed
In your dear little ear
You've got to be carefully taught.
--from South Pacific

Posted by Dave Marain at 5:17 PM 3 comments

Labels: algebra, instructional strategies, math videos, SAT-type problems, video lesson

At the same instant of time, trains A and B enter the opposite ends of a tunnel which is 1/5 mile long. Don't worry -- they are on parallel tracks and no collision occurs!

Train A is traveling at 75 mi/hr and is 1/3 mile long.
Train B is traveling at 100 mi/hr and is 1/4 mile long.

When the rear of train B just emerges from the tunnel, in exactly how many more seconds will it take the rear of train A to emerge?

Click on More to see answer (Feed subscribers should see answer immediately).

Comments
1. Appropriate for middle schoolers even before algebra? Exactly when are middle schoolers in your district introduced to the fundamental Rate_Time_Distance relationship?
2. What benefits do you think result from tackling this kind of exercise? If it's not going to be tested on your standardized tests, is it worth all the time and effort?
3. How much "trackwork" needs to be laid before students are ready for this level of problem-solving?
4. As an instructional strategy, would you have the problem acted out with models in the room or use actual students to represent the trains and the tunnel? OR just have them draw a diagram and go from there? Do a simulation on the TI-Inspire or TI-84 using graphics and parametric equations for the older students?
5. If you believe there is still a place for this type of problem-solving, should it be given only to the advanced classes and depicted as a math contest challenge?
6. I'm dating myself but I remember seeing problems like this in my old yellow Algebra 2 textbook? Uh, I believe this was B.C. -- before calculators! Can you imagine! Do you recall these kinds of problems? Do you recall the author or publisher?
7. Of course, the proverbial "two trains and tunnel" problems are frequently parodied and used as emblematic of the "old math"! They've been replaced by "real-world" applications. "Progress makes perfect!"

YOUR THOUGHTS...





Answer: 9.4 seconds (challenge this if you think I erred!)

Posted by Dave Marain at 6:16 AM 9 comments

Labels: algebra, more, prealgebra, problem-solving, two trains in the tunnel classic, word problems

HAPPY HOLIDAYS!

The following is a series of apparently straightforward arithmetic problems for middle schoolers. However, the objective is to have students justify their reasoning beyond "guess and test" methods. Proving there is only one solution or none requires more careful logic using algebra as needed. Students will need some basic algebra for the "proofs." For the younger student, modify these questions to have them find the squares in questions 1,2,3 and 5. Take this as far as you wish...


In the following, square refers to the square of an integer. Justify your reasoning or prove each of the following.

(1) There is only one square which is 1 more than a prime.

(2) There is only one square which is 4 more than a prime.

(3) There is only one square which is 9 more than a prime.

(4) There is no square which is 16 more than a prime.

(5) There is only one square which is 25 more than a prime.

(6) Can one generalize this or not??

Click Read More for selected answers, solutions...



Selected Answers, Solutions

(2) If n² is 4 more than some prime, p, then we can write
p = n² - 4 = (n-2)(n+2). Since p is prime, the smaller factor must be 1, so
n-2 = 1 or n = 3. Thus, there is only one square, 9, which is 4 more than the prime, 5.

(4) p = n² - 16 = (n-4)(n+4). There n would have to equal 5, n² would equal 25 but 25 - 16 = 9 is not prime.

(6) If there were a general rule would that mean we'd have a formula for primes?

Your thoughts about these questions...

Posted by Dave Marain at 8:04 AM 4 comments

Labels: algebra, middle school, more, number theory, proof

A well-known and intriguing formula usually proved by Mathematical Induction states that
1³ + 2³ + 3³ + ... +n³ = (1+2+3+...+n)² .

In words:
The sum of the cubes of the first n positive integers equals the square of the sum of the first n positive integers (or the square of the nth triangular number).

Students as early as middle school can investigate numerical patterns of sums of powers of positive integers and can be led to such discoveries. However, in this post we will look at a different kind of "proof." Proofs without words can be fascinating, challenging and can develop a student's spatial reasoning. Just as there have been many visual proofs of the Pythagorean Theorem (dissection type), mathematicians have sought visual arguments for many other numerical patterns and algebraic formulas. The Greeks of antiquity developed many classical arguments of this type, necessitated perhaps by not having our symbolic algebra available.

You will surely find other examples of this on the web (e.g., "Cut-the-Knot") but I thought it might be nice to bring it down to a middle school or Algebra 1 level by having students play with some particular cases of this general formula. I have always been intrigued by this topic, ever since I saw several visual proofs of the Pythagorean Theorem. Later on I was introduced to the genius of Sidney Kung and Roger B. Nelson (Google them!). Prof. Kung's extraordinary visual proofs were (and may still be) a staple of Mathematics Magazine, an MAA publication. You may also recall I have published a couple of other such proofs, one of which came from a student of mine. Look here.

Part I
Let's try to demonstrate that 1³ + 2³ = (1 + 2)²

Before displaying the visual we will begin with an arithmetic-algebraic approach:

Think of (1+2)(1+2) as a special case of the form (a+b)(a+b):
Thus, (1+2)(1+2) = (1⋅1) + (1⋅2) + (2⋅1) + (2⋅2)
Now for some creativity. Since cubes involve a product of THREE factors, we can introduce an extra factor of "1" in each term:
(1+2)(1+2) = (1⋅1⋅1) + (1⋅1⋅2 )+ (1⋅2⋅1) + (1⋅2⋅2).

Even without a visual, we can see the first term on the right is 1³!!
It will take some work to show that the sum of the other three terms is 2³. Ok, with this background, here is a
PROOF WITHOUT WORDS
















Do you think your students will "see" the proof?? My crude attempt at a graphic leaves a lot to be desired! It may be helpful to have manipulatives such as algebra tiles available or have students physically build these models. I would encourage that strongly!

So we are proving a numerical formula using a sum of volumes. You might say we turned squares into cubes!!

Do you think this investigation is through? Of course not -- I did all the work for you. Now here is the real test:

Part II

Show that 1³ + 2³ + 3³ = (1 + 2 + 3)²
using a "Proof Without Words."

Ok, I'll give you a little hint although you don't need to use this:
Rewrite (1 + 2 + 3)²
as ((1 + 2 )+ 3)<sup>2

Have fun! Just think, if we have a sum of 4th powers, we might need hypercubes!</sup>

Posted by Dave Marain at 8:20 AM 7 comments

Labels: algebra, patterns, proof without words, spatial sense

Just a 'little' last-minute challenge before Turkey Day -- similar to many you've seen before on this blog and elsewhere...

Determine the exact digits of 100²⁰⁰⁸ - 100¹⁰⁰⁴.


Comments:
Students in middle school or higher will often (or should) employ the "make it simpler and look for a pattern" strategy. Some students will be able to apply algebraic reasoning (factoring, laws of exponents, etc.) to evaluate. It's worth letting students, working in pairs, 'play' with this for awhile, followed by a discussion of various methods. Then challenge them to write their own BIG exponent problem!

Posted by Dave Marain at 6:56 PM 4 comments

Labels: algebra, exponents, math challenge, middle school, patterns

The English language has many confusing phrases but "as many as" IMO has blighted the youth of many an algebra student. Perhaps you think I'm exaggerating this? At the beginning of the school year, write the phrase in the title of this post on the board and have your PreAlgebra/Algebra I (or higher) students write one of the two equations on their paper. Give them only a few seconds, then compile the results. Let us know if the vast majority choose the correct equation. Of course, the outcome depends on the group and many other factors but if we have enough data it might prove interesting. I'm basing this on many years of questioning students. Perhaps I am the only one who has experienced this phenomenon!

The abstraction of algebra is difficult enough for some youngsters. Students who are new to our language have particular difficulty with idiomatic phrases but those born here also seem to struggle with the verbal parts of word problems - that's completely obvious to any algebra teacher of course. If only we could remove the words from a word problem!

Certainly teaching vocabulary and math terminology is an essential part of what we do as instructors. We should also hold students accountable for this vocabulary by assessing it directly.

In this post, I'm inviting readers to share some of the coping mechanisms and pedagogical strategies they use in the classroom to help students survive phrases like "as many as." What phrases seem to cause the most confusion among your students? How about "x is four less than y?"

Here is my initial offering. Let me know if you do something similar or if you feel this might be helpful (or if you vehemently disagree!).

KEY STEP: First decide from the wording of the problem if there are more girls or more boys. In fact, this should have been my original question -- not the equations! It is critical for students to be able to translate the verbal expression into a comparative relationship: Which is the larger quantity? Number of boys or number of girls? Hopefully, most youngsters would interpret the original problem to imply that there are more girls than boys. Hopefully! Ask this question first (metacognitively, students need to learn to ask themselves questions like this when they are reading).

NEXT STEP: Now the issue is where to place the "2" in the equation. Based on the key step above, we know that the number of girls is the larger quantity. Ask them why 2G = B would be incorrect.

Better alternative for some:
We all know that those who have difficulty handling abstraction benefit from concretization, i.e., using numerical values:

Have them write both possibilities:
B = 2G and G = 2B
Now have them substitute values for G and B that make sense for the original problem, say
G = 12, B = 6. Some struggle with this!
By substituting (students like the phrase "plug in") these values into both equations, they should see that 6 = 2⋅12 does not make sense. The correct equation should become apparent. Should...
Of course, most youngsters need to practice many of these before they reach comfort level.

Your thoughts, suggestions, anecdotal evidence???

Posted by Dave Marain at 7:19 AM 16 comments

Labels: algebra, algebra sense, pedagogy, verbal phrases, word problems

Here is an activity for Prealgebra and Algebra students. This introductory activity is not meant to be a conundrum for our crack problem-solvers out there, but the extensions below may prove more challenging.



Target Audience: Grades 6-9 (Prealgebra through Algebra 1)

Major Standards/Objectives:
(1) Representing numerical relationships and patterns algebraically
(2) Recognizing, interpreting and developing function notation
(3) Applying remainder concepts

A 2-column number matrix (grid) is shown above and assumed to continue indefinitely. We will be visiting (traversing) the numbers in the grid starting in the upper left corner with 1. Following the arrows we see that the tour proceeds right, then down, followed by left, then down and repeats.

First, some examples of the function notation we will be using to describe this traversal:
T(1) = 1 denotes that the 1st cell visited contains the number 1.
T(4) = 3 denotes that the 4th cell visited contains the number 3.
Similarly, T(6) = 6.

STUDENT/READER ACTIVITY/INVESTIGATION

(a) Determine T(1), T(5), T(9), T(13), T(17).
(b) 1, 5, 9, 13, 17, ... all leave a remainder of ___ when divided by 4. (Fill in the blank)
Therefore, these numbers can be represented algebraically as 4n + 1, n = 0,1,2,3,...
(c) Based on (a) and (b), it appears that T(4n+1) = _______, where n = 0,1,2,3...
(d) Determine T(2), T(6), T(10), T(14)
(e) 2,6,10,14,... all leave a remainder of ___ when divided by 4. Therefore, these numbers can be represented algebraically as ______, n = _________ (Fill in blanks)
(f) Based on (d) and (e), it appears that T( _____ ) = _____, n = __________.

Note: The instructor may choose to start n from zero or one throughout this activity. I will vary it depending on our needs. It is important for students to see how restrictions (domain of a variable) is critical for an accurate description and that more than one set of restrictions is possible (provided they are equivalent).

Since T(3) = 4 and T(4) = 3, we cannot say that T(n) = n for all n. The numbers 3 and 4 leave remainders of 3 and 0 respectively when divided by 4. We will need a different rule for these kinds of numbers. Let's collect some more data:

(g) By extending the table, determine T(7) and T(8); T(11) and T(12); T(15) and T(16)
(h) Without extending the table, make a conjecture about the values of T(35) and T(36).
(i) Numbers such as 4,8,12,16,... can be represented algebraically as ____, n= 1,2,3,...
(j) Numbers such as 3,7,11,15,... can be represented algebraically as ____, n = 1,2,3,...

Note: Again, the instructor may not like varying the restrictions here. Adjust as needed.

(h) Ok, so you're an expert now. Well, prove it:
T(100) = ______; T(153) = _____; T(999) = ______
Show or explain your method.

EXTENSIONS

Surely, a 3-column number grid or even a 5-column number grid can't be that much more difficult to solve using the same kind of traversal (move to the right until you come to the end, go down, move left until you come to the end, move down, lather, rinse, repeat...). ENJOY!

Ok, for our experts: Try an n x n grid!

DISCLAIMER: As with all of the investigations I publish, these are essentially original creations and therefore have not been proofread or edited by others. You are the 'others!'. You may not only find errors but alternate and perhaps superior ways to present these ideas.
Also, please adhere to the Guidelines for Attribution in the sidebar.

Posted by Dave Marain at 8:07 AM 7 comments

Labels: algebra, functions, investigations, middle school, patterns, prealgebra, remainders

Quick Updates....
Mystery Mathematician Contest ending soon...
Pi fact for today? Try explaining why the imaginary number i raised to the power of i is REAL without mentioning π somewhere! Of course, you could just ask Google to do it for you!


You'd think that the deafening silence from the 5-7-8 triangle post would discourage me - NOT! Here is an investigation for middle schoolers and up.

Typical Content Standard: Patterns, Relations, Algebra

Objectives:
(1) Developing strategies for comparing sums
(2) Developing algebraic generalizations
(3) A few dozen more!

Where might the first question in the title of this post be asked?
(a) SATs?
(b) End of Course Test for Algebra 2?
(c) Other standardized tests?
(d) Math contests? If so, what grade level? 7th? 8th? Higher?

If you value a question such as this, would you introduce it to middle schoolers in 6th? 7th? Prealgebra? Would you use a very different instructional approach with students in higher math courses who have reasonable algebra background? Even if you don't like this question, try it with one of your groups tomorrow and let me know what happens!

Since I have personally posed this type of question to both middle schoolers and older students, I can tell you that even strong math students often have not seen the 'compare differences of corresponding terms method'. I made up that designation but, hopefully, you can make sense of it. Do you think many high school students would attempt to find two separate sums by some method/formula (or using their calculator if allowed) they've seen?

Well, I won't give any more away, but I believe the issues of pedagogy here may transcend the problem and the math strategies:

How does one introduce this? Do you simply have this question on the white board as students enter the room and allow them to work on it individually or in small groups for 5-10 minutes? We all hear about our re-defined role as 'guides on the side' but what exactly does this look like for this activity? How do we facilitate? When do we ask leading questions? What questions would be highly effective here? I haven't even mentioned the calculator issue yet!

So many questions. So few answers...

Actually I was going to do a short video presentation of this question to demonstrate one instructional model, but, unfortunately, my dog ate my main computer which has all my files and applications. Wait - let me apologize to my pooch. He really didn't eat it or even bless it with his bodily functions. But my iBook is very sick and will need intensive care from Apple. In the meantime, I'm on a backup machine, with limited memory and lacking many of my files and applications. Excuses, excuses, excuses! Please bear with me!

Posted by Dave Marain at 10:30 PM 12 comments

Labels: algebra, patterns, sums of consecutive positive integers

Of course this comes a bit late for all those juniors who took their SATs today but the following questions can be used to prepare for the next one OR for anyone who wants to challenge themselves or their students...


-4 ≤ P ≤ 3 and -5 ≤ Q ≤ 4


(i) What is the greatest possible value of (Q+P)(P-Q)? Explain your reasoning.

(ii) What is the least possible value of (Q+P)(P-Q)? Explain your reasoning.

Comments:
(a) SAT questions don't ask for explanations but this goes beyond that.
(b) These would be known as 'grid-ins' or student-constructed response questions. On a real SAT, answers to these must be greater than or equal to zero.
(c) The intent here is to go beyond the typical 'plug-in' methods most students use. One can apply actual skills and reasoning!
(d) Even the strongest students fall into a 'trap' set in these questions. Try them in your classes!
(e) Do these kinds of questions develop algebraic reasoning and mathematical power OR are they just your typical 'tricky' SAT-type that has little value outside the test?
(f) Make up your own version of one of these or, even better, encourage your students to invent their own!

Posted by Dave Marain at 7:36 PM 8 comments

Labels: algebra, algebra 2, SAT-type problems

More of the same...


In a certain group:
The ratio of males to females is 4:5.
The ratio of left-handed people to right-handed is 1:11 (assume no one is ambidextrous!).
64% of of the left-handed people are males.

(a) What % of the males are left-handed?
(b) What % of the females are left-handed?


Comments;

• If students or any of us see enough variations of these, will they become almost mechanical or does one have to decide which method/model/representation is needed for each problem?
• How many models should students be shown for these? Usually students or our readers will find a method or model no one else imagined!
• Is algebra the most powerful method? The most efficient? How many variables? Is it usually best to use one variable and let it represent the total number of people in the group?
  
• Anyone ever use a matrix/spreadsheet/table/Punnett square model to represent the data for these kinds of relationship problems? Specifically, problems in which the entire group (the universe) is divided into either groups A and B or groups C and D. This will become less cryptic as the discussion unfolds.
• Do you think most students today would feel more comfortable working in %, decimal or fraction form? What about rest of us out there?
• Too challenging for middle schoolers or not? Math contest problem or just a challenge to develop facility with ratio thinking? How would most algebra students fare with this?
  

Posted by Dave Marain at 6:21 AM 6 comments

Labels: algebra, middle school math, percent word problem, ratios

Ok, so most normal people are not thinking about the significance of the digit '8' in 2008 the day before the Super Bowl. Sorry, but in this post there will be no predictions about the score, no 'over-unders', no boxes, no betting at all. You do have to admit that this is a great time for lovers of mathematics. People are actually interested in mathematical odds and chances of all kinds of weird number combinations occurring in the score on Sunday night. However, this post will focus instead on the number 8, the units' digit in 2008. The Super Bowl comments above will no doubt soon become outdated but the mathematics below will live on! Who knows, maybe the number 8 will turn out to have special significance on Feb 3, 2008? Remember, I said that here before the game!!

2008 is a special number for so many reasons, being divisible by 4 of course: Leap Year, Prez Election year, Summer Olympics and much more. In fact, 2008 is divisible not only by 4 but also by 8 itself. In the good ol' days, some students were even taught the divisibility rules for 2, 4 and 8:
Divisible by 2: If the 'last' digit is divisible by 2 (of course!)
Divisible by 4: If the number formed by the last TWO digits is divisible by 4
Divisible by 8: If the number formed by the last three digits is divisible by 8.

Let's demonstrate this for 2008:
2008 us divisible by 2 because 8 is divisible by 2
2008 is divisible by 4 because '08' is divisible by 4
2008 is divisible by 8 because '008' is divisible by 8

A little weird with those zeros and not particularly interesting, right? Anyone care to guess a rule for divisibility by 16? Interesting, but none of this is the issue for today....

BACKGROUND FOR PROBLEM/INVESTIGATION/ACTIVITY
Today, we are are interested in the squaresof numbers and their remainders when divided by 8. Notice that 4² is divisible by 8 but 6² is not. So we cannot say that the square of any even number is divisible by 8. What about the squares of odd numbers when divided by 8?
1² leaves a remainder of 1 when divided by 8
3² leaves a remainder of 1 when divided by 8
5² leaves a remainder of 1 when divided by 8
7² leaves a remainder of 1 when divided by 8

What is going on here? That's for your crack investigative team to decipher.

TARGET AUDIENCE: Our readers of course; Middle schoolers through algebra

PROBLEM/INVESTIGATION FOR READERS/STUDENTS
1. Discover, state and prove a general rule for the remainder when the square of an even number is divided by 8.
2. Discover, state and prove a general rule for the remainder when the square of an odd number is divided by 8.

Comments:
(1) These are well-known relationships and not very difficult questions. Just something to extend thinking about divisibility, remainders and the use of algebra to deduce and prove generalizations. Prealgebra students may be able to explain their findings without algebra!
(2) 'Discovering' or stating the rule for question (2) is transparent from the examples above. Instructors may prefer 'data-gathering' and making a table first. That is, have students develop a table for the squares of the first 10 positive integers and their remainders when divided by 8. Proving the result for the squares of odd integers is more challenging, even algebraically. Most will see the remainder when dividing by 4, but 8 is slightly trickier.
(3) Those who are more comfortable with congruences and modular arithmetic can approach these questions another way.

Posted by Dave Marain at 6:28 AM 4 comments

Labels: algebra, divisibility, investigations, proof, remainders

Have you submitted your vote yet in the MathNotations poll in the sidebar?

Target audience for this investigation: Our readers and algebra students (advanced prealgebra students can sometimes find a clever way to solve these).

Let's resurrect for the moment those ever popular rate/time/distance classics. Hang in there -- there's a more interesting purpose here!

We'll start by using fictitious presidential candidates running in a 'race.' Any resemblance to actual candidates is purely coincidental.

R and J are running on a huge circular track. J can run a lap in one month whereas it takes R twelve months to run the same lap. To be nice, J gives R a 3-month head start. After how many months will J 'catch up' to (overtake) R?

Are those of us who were trained to solve these feeling a bit nostalgic? Do you believe that our current generation of students has had the same exposure to these kinds of 'motion' problems or have most of these been relegated to the scrap heap of non-real world problems that serve no useful purpose. Well, they still appear on the SATs, a weak excuse for teaching them, perhaps, but I can certainly see other benefits from solving these. Can you?

Ok, there are many approaches to the problem above. Scroll down a ways to see a couple of methods (don't look at these yet if you want to try it on your own):







Method I: Standard Approach (using chart)

..............RATE ...x.......TIME .....=.....DISTANCE
............(laps/mo).....(months)............(laps)

R.........1/12....................t........................t/12

J................1....................t...........................t

Equation Model (verbal): At the instant when J 'catches up' to R:
Distance (laps) covered by J = Head Start + Distance covered by R

Equation: t = 1/4 + t/12 [Note: The 1/4 comes from the fact that R covers 1/4 of a lap in 3 months]

Solving: 12t = 3 +t --> 11t = 3 --> t = 3/11 months.

Check:
In 3/11 months, J covers 3/11 of a lap.
In the same time, R covers (1/12) (3/11) = 1/44 lap. Adding the extra 1/4 lap, we have 1/4 + 1/44 = 12/44 = 3/11. Check!

[Of course, we all know these fractions would present as much difficulty for students as the setup of the problem, but we won't go there, will we!]

Method II: Relativity Approach
Ever notice when you're zipping along at 65 mph and the car in the next lane is going the same speed, it appears from your vehicle that the other car is not moving, that is, its speed relative to yours is zero! However, if you're traveling at 65 mph and the vehicle in front is going 75 mph, the distance between the 2 cars is ever increasing. In fact, the speedier vehicle will gain 10 miles each hour! This 75-65 calculation is really a vector calculation of course, but, in relativity terms, one can think of it this way:
From the point of view of a passenger in the the slower vehicle, that person is not moving (speed is zero) and the faster vehicle is going 10 miles per hour. We can say the relative speeds are 0 and 10 mph.

Ok, let's apply that to the 'race':

If R's relative speed is regarded as zero, then J's relative speed will be 1 - 1/12 = 11/12 laps/month.
Since R is not 'moving', J only needs to cover the head-start distance to catch up:
(11/12)t = 1/4 --> t = (1/4)(12/11) = 3/11 months. Check!
[Note: Like any higher-order abstract approach, some students will latch on to this immediately and others will have that glazed look in their eyes. It may take some time for the ideas to set in. This method is just an option...]

There are other methods one could devise, particularly if we change the units (e.g., working in degrees rather than laps). Have you figured out how all of this will be related to those famous clock problems? Helping students make connections is not an easy task. One has to plan for this as opposed to hoping it will happen fortuitously.

Here is the analogous problem for clocks:

At exactly what time between 3:00 and 4:00, will the hour and minute hands of a clock be together?

Notes:
(1) I will not post an answer or solution at this time. I'm sure the correct answers and alternate methods will soon appear in the comments.
(2) A single problem like this does not an investigation make. How might one extend or generalize this question? Again, these are well-known problems and I'm sure many of you have seen numerous variations on clock problems. Share your favorites!
(3) Isn't it nice that analog watches have come back into fashion so we can recycle these wonderful word problems!
(4) For many problem-solvers, part of the difficulty with clock problems is deciding what units to use for distance (rotations, minute-spaces, some measure of arc length, degrees, etc.). This is a critical issue and some time is needed to explore different choices here.

Posted by Dave Marain at 6:20 AM 6 comments

Labels: algebra, clock problems, percent word problem, rate-time-distance problems

Posted by Dave Marain at 5:58 AM 2 comments

Labels: algebra, factoring

Update:
(1) See the visualization for the difference of squares posted on 1-9-08.
(2) Read the comments in this post for considerable clarification and instructor guidelines and suggestions. Mathmom's and Eric's comments are particularly insightful.

This post can be developed into an activity for prealgebra through first-year algebra students (or even 2nd year algebra). The last part is more challenging.
The focus here is on developing a method/strategy that can be used to solve similar Diophantine equations. The other objective is to introduce the ideas and methods of proof. This problem may later be used to solve a recent math contest problem for which I obtained permission to discuss on this blog. I am fully aware that many students will 'solve' these equations by Guess-Test methods, but they need to go further.

STUDENT/READER PROBLEMS/ACTIVITIES

(a) Prove there is only one solution in positive integers for the equation:

M² - N² = 12

Note: If we omit the word positive, what would the solution(s) be?

(b) Determine all positive integer solutions:

M² - N² = 15

(c) Determine all positive integer solutions:

M² - N² = 36

(d) Let's investigate for what positive integer values of P, M² - N² = P has NO solutions in positive integers.

(i) Determine at least 5 positive integer values of P for which the above equation has no positive integer solutions.

(ii) (More challenging) Describe all values of P for which the above equation has no solutions. Justify your result.

Note: All students should have success with (i), although some may struggle to find 5 values. Part(ii) should challenge the student who has finished the other parts in rapid order and sits there complacently!

Additional Comment: If P is itself a perfect square, our equation is obviously related to the most famous equation in geometry. Thus, if P = 9 or P = 16, for example, students should recognize something! For this reason you may want to have students consider these values when doing this investigation. More to come...

Posted by Dave Marain at 4:00 PM 8 comments

Labels: algebra, algebra 2, Diophantine equation, middle school