Kiss those Math Headaches GOODBYE!

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Everyday Life Sparks Mathematical Puzzles


So here’s the situation: you’re at the breakfast table, enjoying a bowl of steaming-hot steel-cut oats and maple syrup, and you just poured yourself a mug of black coffee. But then you realize you want to pour some milk in the coffee (sorry, purists). But the milk is in the frig, six feet away. So of course you walk to the frig, grab the milk, bring it to the table, pour some in your coffee, return the milk to the frig and sit back down. Question: could you have done this more efficiently?

Yes, of course. You could have brought your cup of coffee with you as you walked to the frig, poured the milk right there at the frig, returned the milk, and then walked back to the table.

Being Smart?

“Morning Joe”

When I realized this this morning, I thought … hmmm. Had I used a bit of forethought, I would save myself an entire round trip from the table to the frig. And while I have no problem making that extra trip (hey, just burned 1.3 calories, right?), the experience made me wonder if anyone has ever developed a mathematics of efficiency for running errands.

I could imagine someone taking initial steps for this. One would create symbols for the various aspects of errands. There would be a general symbol for an errand, and there would be a special ways of denoting: 1) an errand station (like the frig), 2)  an errand that requires transporting an item (like carrying the mug), 3) an errand that requires doing an activity (pouring milk) with two items (mug and milk) at an errand station, 4) an errand that involves picking something up (picking up the mug), and so on. Then one could schematize the process and use it to code various kinds of errands. Eventually, perhaps, one could use such a system to analyze the most efficient way to, say, carry out 15 errands of which 3 involve transporting items, 7 involve picking things up, and 5 involve doing tasks at errand stations. Don’t get me wrong! I have not even begun to try this, but I’ve studied enough math that I can imagine it being done, and that’s one thing I love about math; it allows us to create general systems for analyzing real-world situations and thereby to do those activities more intelligently.

Of course, one reason I’m bringing this up is to encourage people to think more deeply about things that occur in their everyday lives. Activities that appear mundane can become mathematically intriguing when investigated. A wonderful example is the famous problem of the “Bridges of Konigsberg,” explored by the prolific mathematician Leonhard Euler nearly 300 years ago.

Euler in 1736 was living in the town of Konigsberg, now part of Russia. The Pregel River, which flows through Konigsberg, weaves around two islands that are part of the town, and a set of seven lovely bridges connect the islands to each other and to the town’s two river banks. For centuries Konigsberg’s residents wondered if there was a way to take a walk, starting at Point A, crossing each bridge exactly once, and return to Point A. But no one had found a way to do this.

One of the famous Seven Bridges of Konigsberg

One of the famous Seven Bridges of Konigsberg

Enter Euler. The great mathematician sat down and simplified the problem, turning the bridges into abstract line segments and transforming the bridge entrance and exits into points. Eventually Euler rigorously proved that there is no way to take the walk that people had wondered about. This would be just an interesting little tale, but it has a remarkable offshoot. After Euler published his proof, mathematicians took his way of simplifying the situation and, by exploring it, developed two new branches of math:  topology and graph theory. The graph theory ideas that Euler first explored when thinking about the seven bridges sparked a branch of math that’s used today to determine the most efficient ways of connecting servers that form the backbone of the internet!

Of course, there’s also the classic example of Archimedes shouting “Eureka!” and running through the streets naked after seeing water rise in his bathtub. In that moment, Archimedes, who had been trying to help his king figure out if the crown that was just made for him had been created with pure gold, or with an alloy, saw that the water displacement would help him solve the problem. In the end, Archimedes determined that the crown was not pure gold, and the king rewarded the great thinker for his efforts.

As I write this, I find myself wondering if any of you readers can think of other situations in which everyday life experiences led mathematicians or scientists to major discoveries. It would be enlightening to hear more of these stories.

And, if no such stories spring to mind, check out this site, which lists several such stories.  http://www.sciencechannel.com/famous-scientists-discoveries/10-eureka-moments.htm

In any case, the way that such discoveries occur shows that you never know where a seemingly trivial idea might lead … so it’s good to keep your eyes and mind open.

How to Find the GCF for Three or More Numbers


To find the GCF for three or more numbers,  follow these steps:

1)  Determine which of the given numbers is smallest, then find the smallest difference between any pair of numbers.

2)  See what is smaller:  the smallest number, or the smallest difference. Whichever one  is smallest, that number is the GPGCF (Greatest Possible GCF). That means that this is the biggest number that the GCF could possibly be. Or, more formally we would say:  The GCF, if it exists, must be less than or equal to the GPGCF.

3)  Check if the GPGCF itself goes into all of the given numbers. If so, then it is the GCF. If not, list the factors of the GPGCF from  largest to the smallest and test them until you find the largest one that does divide evenly into the given numbers. The first factor (i.e., the largest factor) that divides evenly into the given numbers is, by definition, the GCF.

EXAMPLE:

Problem:  Find the GCF for 18, 30,  54.

1)  Note that the smallest number is 18, and  the smallest difference between the pairs is 12 [54 – 30 = 24;  54 – 18 = 36;  30 – 18 = 12] .

2)  Of those four quantities (the smallest number and the three differences), 12 is the least. This means that the
GPGCF = 12.

3) Check if 12 divides evenly into the three given numbers: 18, 30 and 54. In fact, 12 doesn’t divide evenly into ANY of these  numbers. Next we check the factors of 12, in order from largest to smallest. Those factors are: 6, 4, 3 and 2. The first of those that divides evenly into all three numbers is 6. [18 ÷ 6 = 3;  30 ÷ 6 = 5;  54 ÷ 6 = 9]. So the GCF = 6. And we are done.
MORE CHALLENGING PROBLEM:

Find the GCF for 24, 148, 200.

1)  Note that the smallest number is 24, and that the smallest difference between the pairs is 52 [200 – 148 = 52;  200 – 24 = 176;  148 – 24 = 124] .

2)  Of those four quantities (the smallest number and the three differences), 24 is the least. This means that for this problem, the GPGCF = 24.

3) Check if 24 divides evenly into the three given numbers: 24, 148 and 200. While 24 does divide evenly into 24, it does not divide evenly into 148 or 200. So next we check the factors of 24, in order from largest to smallest. Those factors are: 12, 8, 6, 4, 3 and 2. The first of those that divides evenly into the three given numbers is 4. [24 ÷ 4 = 6;  148 ÷ 4 = 37;  200 ÷ 4 = 50]. So the GCF = 4. And, once again, we are done.

The process may seem a bit long, but once you get used to it and start doing it in your mind, not on paper, you should find that it actually is quite fast. And you’ll find yourself figuring out the GCF for three or more numbers all in your mind — with no need for pencil and paper — while everyone around you will be making prime factor trees or using calculators. And surely that is a good feeling.

Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like how Josh explains these problems, you’ll certainly  like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com  Just click the links in the sidebar for more information! 

How to Combine Positive & Negative Numbers — Quickly and Easily


If you or someone you know struggles when combining numbers with opposite signs — one positive, the other negative — this post is for you!

To be clear, I’m referring to problems like these:

 – 2 + 7 [first number negative, second number positive], or

+ 13 – 20 [first number positive, second number negative]

To work out the answers, turn each problem into a math-story. In this case, turn it into the story of a tug-of-war battle. Here’s how.

In the first problem, – 2 + 7, view the – 2 as meaning there are 2 people on the “negative” team; similarly, view the + 7 as meaning there are 7 people on the “positive” team.

There are just three things to keep in mind for this math-story:

1)  Every “person” participating in the tug-of-war is equally strong.

2)  The team with more people always wins; the team with fewer people always loses.

3)  In the story we figure out by how many people the winning team “outnumbers” the other team. That’s simple; it just means how many more people are on that team than are on the other team. Example: if the negative team has 2 people and the positive team has 7 people, we say the positive team “outnumbers” the negative team by 5 people, since 7 is 5 more than 2.

Now to simplify such a problem, just answer three simple questions: 

1)  How many people are on each team?
In our first problem, – 2 + 7, there are 2 people on the negative team and 7 people on the positive team.

2)  Which team WINS?
Since there are more people on the positive team, the positive team wins.

3) By how many people does the winning team OUTNUMBER the losing team?
Since the positives have 7 while the negatives have only 2, the positives outnumber the negatives by 5.

Now ignore the answer to the intro question, Question 1, but put together your answers to Questions 2 and 3.

ANSWER TO QUESTION 2:  +

ANSWER TO QUESTION 3:  5

ANSWERS TOGETHER:  + 5

All in all, this tells us that:  – 2 + 7 = + 5

For those of you who’ve torn your hair out over such problems, I have good news …

… THEY REALLY ARE THIS SIMPLE!

But to believe this, it will help to work out one more problem:  + 13 – 20.

Here, again, are the common-sense questions, along with their answers.

1)  How many people are on each team?
In this problem, + 13 – 20, there are 13 people on the positive team and 20 people on the negative team.

2)  Which team WINS?
Since there are more people on the negative team in this problem, the negative team wins.

3) By how many people does the winning team OUTNUMBER the losing team?
Since the negatives have 20 while the positives have only 13, the negatives outnumber the positives by 7.

Just as you did in the first problem, put together your answers to Questions 2 and 3.

ANSWER TO QUESTION 2:  

ANSWER TO QUESTION 3:  7

ANSWERS TOGETHER:  – 7

All in all, this tells us that:  + 13 – 20  = – 7

Now try these for practice:

a)  – 3 + 9

b) + 1 – 4

c)  –  9 + 23

d)  – 37 + 19

e) + 49 – 82

Answer to Practice Problems:

a)  – 3 + 9 = + 6

b) + 1 – 4 = – 3

c)  –  9 + 23 = + 14

d)  – 37 + 19 = – 18

e) + 49 – 82 = – 33

Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like the way Josh explains these problems, you will very likely like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com  Just click the links in the sidebar for more information! 

How to Decrease Mistakes in Algebra – Part 3


When we left off, we were talking about the double-slash, a form of notation I’ve developed that helps students attain greater focus when simplifying algebraic expressions.

With greater focus, students make fewer mistakes. With the double-slash at their disposal, students avoid the mistake of combining terms that should not be combined. In the following example, students use the double-slash twice to simplify an algebraic expression:

     + 8 – 2(3x – 7)

=            + 8   //  – 2(3x – 7)

=            + 8  //  – 6x + 14

=            – 6x  //  + 8 + 14

=            – 6x  + 22

No Mistakes

Let's Reduce Mistakes in Algebra!

By cordoning off the section with the distributive property:  – 2(3x – 7), the double-slash allows students to see it distraction-free. With this heightened level of focus, students are more likely to work out the distributive property correctly, then continue on, simplifying the whole expression with no mistakes.

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“Math Cafe” Gives Students Options for Math Success


Sometimes when I tutor I tell students that they are “hanging the in Math Cafe.”

I explain that when I tutor, I like to offer students a “menu of math options.”

Math Cafe

Math Cafe, Open 24/7 = 3.42857 ...

Instead of showing students just one way to work a math problem, I try to present a “menu” of approaches, and I tell students that they need to listen so they can decide which approach works best for them.

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