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Archive for the ‘Number Theory’ Category

How to Let Kids Use Calculators Without Ruining Their Ability to Think Numerically


Hi folks,

Now that summer has officially begun, I’m enjoying a certain distance from the heat of the school year, and that distance gives me a chance to reflect.

One set of ideas that my mind keeps poking around again and again is this:  a) the weakness in actual number sense among today’s elementary and secondary students,
b) the concomitant modern focus on teaching Number Sense during these school years, and c) the now-rampant overuse of calculators.

I find it interesting that Number Sense has become a “big important new topic” that math instructors are required to teach. I also find it interesting that the new focus on Number Sense has been growing steadily at the very same time that students in so many parts of our country have become more and more calculator dependent.

Could there be a connection?

Yes, undoubtedly! Back when I set up shop tutoring math, K-12, in 1990, Santa Fe (NM) Public School students were not permitted to use calculators willy-nilly. Because of that, our students were not calculator-dependent. Students were expected to know the truths of arithmetic forwards and backwards, and wouldn’t have dreamed of reaching for a calculator to find the value of something so simple as, say, 7 + 5, as happens routinely today. Yes, routinely! I should know; I’m a professional math tutor.

What’s more, I’d say that students in the 1990s generally understood concepts such as odd and even numbers, prime and composite numbers, how to prime factorize, how to find the GCF and the LCM, and the many other skills that are part of the “new area of math instruction we call Number Sense.

That’s because teachers used to require students to use their minds to work with numbers. Students used to grind out 7/18 + 5/12 by hand, not by pressing buttons. They used to figure out the LCM of 22 and 30 by using an algorithm rather than by tapping an app. They used to prime factorize numbers using the good old factor tree and simplify radicals by thinking rather than by pressing a sequence of buttons and scrolling through the numbers flashing across their LCDs.

You can probably see where I’m going with this. Today’s math students have become overly calculator dependent. That dependence on calculators, in turn, has made them deficient at the skills in the topic area we call Number Sense. And precisely because today’s students are so deficient at number sense, precisely because they have been allowed to become so dependent on their e-devices rather than on their mental devices, curriculum designers have devised this whole new area of math, Number Sense, that now gets taught as its own “thing” rather than being an integral thread of everyday math instruction. Number sense used to be something students developed naturally, by mentally working with numbers, day-in, day-out, using paper and pencil and mental math.

Lest I be called a Luddite, I’m not saying that calculators have no place in the math curriculum. But as a tutor who has helped students with math for some 27 years now, I can say with certainty that today’s students’ innate ability to work with numbers, play with numbers and calculate with numbers has been dulled and frankly allowed to atrophy because calculators have become an all-too-easy, all-too-available crutch.

In this way, math curricula and math educators who overly promote calculator usage have done a great disservice to students. The good news, though, is that  teachers could correct course without too much trouble.

Teachers could still allow students to use calculators, quite appropriately, for higher-order processes — such as graphing two functions to see where they intersect, and to see if the answer found that way comports with the answer attained by solving the systems simultaneously by hand — while at the same time disallowing calculator usage for arithmetic calculations.

I’d like to see teachers get their students back to basics in this way because, from my perspective, we’re raising a new generation of students, many of whom have little ability to calculate mentally and little understanding of how numbers work. As a result, these children (soon-to-be adults) are unnecessarily vulnerable.

They’re vulnerable because they cannot tell if they are receiving the correct change from a cashier. They’re vulnerable because they cannot tell if their car or home interest payment is correct. And they’re vulnerable in a larger sense because they lack the ability to easily think numerically, i.e., quantitatively. And when people lack the fundamental ability to think quantitatively, even having a calculator won’t save them in many situations. That’s because they might not even know what operation to do to find a solution in a real-world situation.

But in an even more direct and practical sense, the new calculator-dependent students are vulnerable because they have been set up to struggle mightily in their college math classes. That’s because nearly all U.S. colleges require students to take math tests without using calculators!

So I say let’s get back to basics, and let’s do it in a smart way. Let’s continue to let students use calculators for higher-order thinking skills, but let’s disallow calculators for ALL arithmetic so that students will be required to once again become strong in those critical fundamental skills and so that they will re-gain the natural form of Number Sense that is their right and their due.

 

 

 

 

 

Math Trick: Find the GCF for 3 or More Numbers


Find the GCF, your teacher says … not just for 2 numbers, but for 5 of them.

And yes, you need to do it by prime factorizing.

Can’t you just hear the students’ groans?!

But what if there were a way to do this without prime factorizing? Could it really be?

Yes!

What I’m about to teach you is a technique that lets you find the GCF of as many numbers as you wish, and with much greater ease than the old factoring technique. (by the way, I don’t really hate the factoring technique … it actually teaches you a lot about numbers … but it can get annoying!).

So why don’t they teach this new way in school? No idea. But let’s just focus on the technique because once you do, you’ll be so much faster at finding the GCF …  you’ll be amazing your friends and your teacher, too!

So just kick back, watch the video — and learn …. then do the practice problems at the end of the video, to become a whiz! And remember, if you ever want extra help in the form of tutoring, I’m available — worldwide — thanks to the power of online videoconferencing.

Enjoy!

— Josh

 

 

 

How to find the GCF of 3+ Numbers — FAST … no prime factorizing


Suppose you need to find the GCF of three or more numbers, and you’d really prefer to avoid prime factorizing. Is there a way? Sure there is … here’s how.

 

High-Octane Boost for Math

High-Octane Boost for Math Ed

Example:  Find the GCF for  18, 42 and 96

Step 1)  Write the numbers down from left to right, like this:

………. 18     42     96

[FYI, the periods: …. are there just to indent the numbers. They have no mathematical meaning.]

Step 2)  Find any number that goes into all three numbers. You don’t need to choose the largest such number. Suppose we use the number 2. Write that number to the left of the three numbers. Then divide all three numbers by 2 and write the results below the numbers like this:

2    |  18     42     96
……..  9     21     48

Step 3)  Find another number that goes into all three remaining numbers. It could be the same number. If it is, use that. If not, use any other number that goes into the remaining numbers. In this example, 3 goes into all of them. So write down the 3 to the left and once again show the results of dividing, like this:

2    |  18     42     96
3    |    9     21     48
……… 3      7      16

Step 4)  You’ll eventually reach a stage at which there’s no other number that goes into all of the remaining numbers. Once at that stage, just multiply the numbers in the far-left column, the numbers you pulled out. In this case, those are the numbers:  2 and 3. Just multiply those numbers together, and that’s the GCF. So in this example, the GCF is 2 x 3 = 6, and that’s all there is to it.

Now try this yourself by doing these problems. Answers are below.

a)   18, 45, 108
b)   48, 80, 112
c)   32, 72, 112
d)   24, 60, 84, 132
e)   28,  42, 70, 126, 154

Answers:
a)   GCF =  9
b)   GCF =  16
c)   GCF =  8
d)   GCF =  12
e)   GCF =  14

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! 

The “Ladder of Primes”


Remember the best teachers you had? Remember how they made their classes come alive? How one of the ways they made things exciting was by using analogies — little stories that connected new concepts to things you already knew and understood?

Educational researchers today are studying what makes analogies such an effective teaching tool. They have found that the use of analogies is one of the best techniques for making concepts “stick.” By relating that which students need to know to that which they already do know, teachers create bridges in understanding, and those bridges give students a way to grasp a new and difficult concepts.

The same holds true in math class. If we teachers use powerful analogies that make concepts more memorable, students are more likely to enjoy the lesson, and as a result, they’ll be more likely to remember what was taught.

I would like to present a quick-and-easy analogy that helps students learn about our number system, on the one hand, and which also helps students work with fractions, on the other hand.

The analogy is to something I call the “Ladder of Primes.” (more…)

How to Understand the LCM (Part 1)


I don’t know about you folks, but I’ve always been a bit disappointed by the various techniques for finding the Least Common Multiple (LCM) for a pair of numbers.

While there are several techniques that “work” — by which I mean techniques we can teach to students and have them learn quickly — I’ve known of no technique that makes good intuitive sense. In other words, I’ve known no technique whose underlying principle felt obvious.

Feeling frustrated, I started looking for a technique that would have that undeniable “ring of truth.”

Coffee, Pi and More

Coffee, Pi and More

And so, after playing around in my “sandbox of numbers” for quite a while,  I’m happy to report that I’ve finally found what I had been looking for.

In today’s post I will show you a way to find the least common multiple that makes sense, at least to me. I hope it will make sense to you as well.

(more…)

Quick Easy Way to Untangle Confusion re: “Greater” and “Less”


Whenever I can find a memory trick that helps students get something straight, I use it. Students needs to remember so many things in algebra, so whatever help we can give them is well appreciated.

So recently I stumbled upon a memory trick that helps students tell which of two numbers is greater and which is less.

No Mistakes

Let's Reduce Mistakes in Algebra!

You might be thinking:  greater and less?! Why would any student have trouble with that? Well, before students hit negative numbers and absolute value, there is generally little trouble. The greater numbers are the larger numbers, the lesser numbers are the smaller numbers. And kids basically know what we mean by larger and smaller whole numbers, when they are dealing with positive numbers and zero.

But when students encounter negative numbers, some things change.
While 10 > 5,   – 10 IS not > – 5. Instead:  – 10 < – 5.

As if that were not enough, absolute vale comes along and makes things still more confusing, since it takes the value of any number and makes it positive. So now:

abs. value of – 10 > abs. value of – 5

(more…)

Secrets of the Calculator, Part 2


Welcome to the second in my series of posts about the “secrets of the calculator.” In my first post on this topic I explained how the 1/x key can make it easier to input certain expressions. If you missed that one, just click here.

Today I’d like to talk about another key that is quite useful, the “EE” key.

What in the world is the EE key, and what does it do?

At first hearing, EE sounds a bit like the call of a monkey.

(more…)

How to Use the Calculator’s 1/x Key?


In my tutoring I am continually surprised by how little most students know about their calculators.

It is true that most students know the basics:  the four operations, the exponent key, the square root key, the Pi key, and maybe some trig fundamentals, like sin, cos 60 and tan. But aside from these basic keys and keystrokes, many students have little to no idea what the other keys do.

The funny thing is that there are so many keys and keystrokes that students would just “love” if they only knew about them.

So to help students out a bit here, I’m starting an occasional series whose name is just below.

(more…)

How to See Why the Divisibility Trick for 3 Works


One of my subscribers asked why the trick for divisibility for 3 actually works. [If you missed the post on that trick, go here:]

But the gist of the trick is this:  3 divides evenly into a number if 3 divides evenly into the sum of the digits of the number.

I’ll prove this trick for a three-digit number, but you’ll see why the proof applies to numbers with as many digits as you’d like.

Let’s call our three-digit number cde, where c is the digit in the 100s place, d is the digit in the 10s place, and e is the digit in the 1s place.

We can state the value of our cde number like this:

cde =  (100 x c)  +  (10 x d)  +  e

This shows that the c-part of the number is made up of 100 groups of c. If you think about it, there’s no reason we can’t re-write the value of this digit as 99 groups of c plus 1c, or just:  (99c + c).

In the same way, the d-part of cde is made up of 10 groups of d, which we can re-write as (9d + d). And of course the e-part of the number is just e.

So now we have this:

cde =  (99c +  c)  +  (9d + d)  +  e

Using the rules of jolly old algebra, we can shuffle the terms around a bit to get this:

cde =   99c + 9d + (c + d + e)

Then, adding a set of parentheses for clarity, we get this:

cde =   (99c + 9d) + (cde)

So far, so good. But what’s the point? Well, we’re just getting to that.

Let’s think a bit more about the 99c term. Factoring out a 3, we see that 99c =  3 x 33c. Since 3 is a factor of this expression, 99c must be divisible by 3. Aha, progress, right?

Factoring the d-term, we see that 9d = 3 x 3d. Again, since 3 is a factor of this expression, 9d is  also divisible by 3.

So we now know that both 99c and 9d are divisible by 3.

We can never forget the Divisibility Principle of Sums (DPS), which says:

If a number, x, divides evenly into both a and b, then it divides evenly into their sum, (a + b).

What does that mean here? It means that since 3 divides evenly into both 99c and 9d, it must divide evenly into their sum:   (99c + 9d).

And remember that our number, cde, equals nothing more than:  (99c + 9d) + (cde)

So, we’ve just found out that 3 divides evenly into the quantity in the first parentheses:  99c + 9d

So to find out if 3 divides evenly into the whole number cde, all that’s left is to find out whether or not 3 divides into what remains, the quantity in the second parentheses:  c + d + e

But guess what? c + d + e is the sum of the digits for our number, cde. So this idea right here is the trick for divisibility for 3:  To find out if 3 divides evenly into a number, just add up the number’s digits and see if 3 goes into that sum.  If it does, then 3 does go in; if not, 3 does not go in.

So this proves the divisibility trick for a three-digit number like cde.

To see why the same trick works for numbers with four or more digits, keep in mind that the larger digits can similarly all be broken up as we broke up the digits of c and d. For example, if we have a four-digit number, bcde, then the value of the leading digit b can be viewed, first, as (1,000 x b). And then this can be split apart once more, into 999b + b. That way, this four-digit number fits into the pattern of the trick. And this same kind of split up can be done for any digit whatsoever.

So if you follow the logic of this proof, you now see that the divisibility trick rests on solid logical/mathematical ground.