Kiss those Math Headaches GOODBYE!

Archive for November, 2008

Tutor Tales #1: Color in Geometry


This if the first in what I hope will be a long series of brief blogs called Tutor Tales blogs.

The idea is that while I’m tutoring I get ideas or insights on how to help students, and then I write up a short blog entry on that experience, preferably on the day that the event occurred.

I hope that these Tutor Tales will give you examples of approaches to math that help students (or that do not help, depending on what I did), and that they give you a chance to reflect on your own teaching.

For the first Tutor Tale entry, I just noticed how useful it can be to use color in geometry.

The girl I was tutoring had a problem:  Find out how many diagonals can be drawn inside a regular, convex nine-sided polygon.

I’ve already noticed that this girl likes color, and she is 17 years old. So I had a hunch that she would be open to trying a color-approach.

We created the non-agon by first drawing a circle, and then marking off nine points on the circle. Then we connected the points sequentially.

To find out how many diagonals we could draw for such a figure, we chose one color for the top point, green, and drew all of the diagonals we could for that point, in green. It turned out that there were 6 diagonals, so we put a big 6 in green at this vertex. Then we tried the next vertex, which we colored pink. We found that we could create 6 additional diagonals from this vertex, and we colored these pink. So we put a big pink 6 by this vertex. We went around the circle in a clockwise way, using a different color for each vertex. All in all we found that the pattern of diagonals was:  6, 6, 5, 4, 3, 2, 1, 0, 0, for a grand total of 27 diagonals.

Here’s the image of the figure we worked on.


One thing to consider, especially if you teach geometry, is how many opportunities there are in geometry to use color to separate different concepts and to relate similar concepts. Check it out and see what you discover.

Happy Teaching!

—  Josh

How to Practice Dividing Fractions

Last week I sent out a blog describing my FRACTION SANDWICH approach to dividing a fraction by a fraction. The title of the blog was: Dividing Fractions:  From Annoying to FUN!

In today’s post I offer ten solid problems that help students practice the FRACTION SANDWICH, with answers just below.

Note:  if your students struggle with these problems, it’s very possible that some of them could brush up on the rules of divisibility. For a quick review of those rules, go to:

Try these:

a)  (2/9) / (8/12)

b)  (12/16) / (15/40)

c)  (36/21) / 63/35)

d)  (44/24) / (77/48)

e)  (15/49) / (21/56)

f)  (39/52) / (24/56)

g)  (27/45) / (33/65)

h)  (12/18) / (28/45)

i)  (17/51) / (99/121)

j)  (28/18) / (42/54)


a)  (2/9) / (8/12)  =  1/3

b)  (12/16) / (15/40)  = 2

c)  (36/21) / 63/35)  =  20/21

d)  (44/24) / (77/48)  =  1 and 1/7

e)  (15/49) / (21/56)  =  40/49

f)  (39/52) / (24/56)  =  1 and 3/4

g)  (27/45) / (33/65)  =  1 and 2/11

h)  (12/18) / (28/45)  =  1 and 1/14

i)  (17/51) / (99/121)  =  1/27

j)  (28/18) / (42/54)  =   2

Multiplication Trick #5 — How to Multiply Two-Digit Numbers by 11

This is the fifth in my series on multiplication tricks. I suggest that you make mental math “tricks” a steady part of your math instruction. Benefits students will reap include:

—  delight with the tricks themselves

—  enhanced confidence in working with numbers

—  students who otherwise don’t like math — or don’t like it much — often find the tricks irresistibly fun and interesting


WHAT THE TRICK LETS YOU DO: Multiply two-digit numbers by 11.

HOW YOU DO IT:  To multiply a two-digit number by 11, first realize that the answer will have three digits. The first (left-most) digit of the answer is the first digit of the number; the last (right-most) digit of the answer is the last digit of the number; and the middle digit is the sum of the first and last digits.

But those are just words … here’s a living, breathing example …

Example:  11 x 25


Look at 25. The first digit is 2; the last digit is 5.

First digit of answer is 2, so thus far we know the answer looks like:  2 _ _

Last digit of answer is 5, so now we know the answer looks like:  2 _ 5

Middle digit is 7, since 2 + 5 = 7.

The answer is the three-digit number:  2 7 5, more casually known as 275.

It’s that easy!


First digit of answer is 6, so thus far we know the answer looks like:  6 _ _

Last digit of answer is 3, so now we know the answer looks like:  6 _ 3

Middle digit is 9, since 6 + 3 = 9.

The answer is the three-digit number: 6 9 3, or just 693.

Try these for practice:

11 x 24

11 x 31

11 x 52

11 x 27

11 x 34

11 x 26

11 x 62


11 x 24 = 264

11 x 31 = 341

11 x 52 = 572

11 x 27 = 297

11 x 34 = 374

11 x 26 = 286

11 x 62 = 682

NOTE:  If you’re clever (and we’re sure that you are), you have probably realized that this trick, as described, works only when the digits add up to 9 or less. So what do you do when the digits add up to 10 or more? Some of you may figure this out on your own. For those who need a little help, the answer to this will be included in an upcoming blog post.

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  Just click the links in the sidebar for more information! 

How to Divide Fractions: from annoying to FUN!

O.K., I’m ready to share my amazing approach to dividing a fraction by another fraction. Well, maybe not breathtaking … like Andrew Wiles’ proof of Fermat’s Last Theorem … but at least interesting. And best of all, fun and student-friendly!

Last week I asked if anyone had any tricks up their sleeves that make it easier for students to divide fractions. And I said that I would share a trick after I heard from you.

I got a nice response from Michelle, who said that she has used the mnemonic “KFC” (like the fried chicken), which in her class stands for Keep-Change-Flip. The idea being that you KEEP the first fraction, and next you CHANGE the sign from multiplication to division. Finally you FLIP the second fraction, the fraction on the right. We have similar mnemonic where I live, which goes by the phrase: Copy-Dot-Flip, with the “dot” meaning the dot of multiplication.

But what I want to share with you is a completely different approach to dividing one fraction by another, an approach that saves time, and makes it both easier and more fun — in my humble opinion — than the standard approach.

The approach I’m going to show you works for any complex fraction situation you might encounter, such as these:

For this blog post, I’m going to limit my chat to complex fractions of the arithmetic type, meaning those with numbers only, and no variables. And if it seems important, I’ll do another post later on using this very same process for algebraic fractions.

So what is this amazing approach, anyway? Well, it’s based on something I discovered on day when I was just messing around with fractions divided by fractions. I realized that after you do the KFC or the Copy-Dot-Flip, what you get — in general — is actually something really easy to grasp, as this next image will show you, along with a Quick Proof:


If you take a moment to think about it, the terms in the numerator of the result — terms a and d — have something in common; they were on the outside of the original complex fraction, so I call these terms the “outers.” In the same way, the terms in the denominator of the result — terms b and c — were both on the inside of the complex fraction, so I call them the “inners.”

So when you divide fractions in this vertical format, the answer is simply the outers, multiplying each other divided by the inners, multiplying each other.

I find that students find this easy to remember and a cinch to do. This next sheet summarizes the idea, and also provides a fun way of remembering the concept, thinking about the stack of terms as a fraction “sandwich.”


So, to put this in words, the four-level complex fraction that you start out with can be thought of as a sandwich, with two pieces of bread at top and bottom, and slices of bologna and cheese in the middle.

The main point is that to simplify the fraction sandwich, all you need to do is put the two slices of bread together in the numerator and multiply them, And then put the bologna and cheese together in the denominator, and multiply them.

Using this idea it becomes a lot easier to simplify these complex fractions. Here’s an image that shows how it is done, and how this approach saves time over the way we were taught to do it, using reciprocals.


And there’s more good news. This new way of looking at complex fractions also gives students a cool, new way to simplify the fractions before they get the answer. And when you do simplify fully, the answer you get will be a fraction that’s already completely reduced, so you won’t have to stress about that part.

The next two pages show you this fun and easy new way to simplify:


or, or what? …  Here’s what …


So now you might like to see the whole process from start to finish, so you can decide for yourself if this technique is for you. Well that’s exactly what we’re showing next. As you can see I consistently highlight the outers with pink, the inners with yellow.


And finally, a “harder” problem, you might say. But check it out. Is it really any harder than the one we’ve just done? You decide.


In my next blog I’ll give you a few problems like these, so you can get used to this trick, and start shaving precious seconds and nano-seconds off the time it take you to do your homework, so you spend more time doing all of those things that you want to do more:  texting, watching You-Tube, taking hikes, skating (roller and ice), etc. etc. , etc. You know better than me.

Happy Teaching and Learning!

—  Josh


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  Just click the links in the sidebar for more information! 

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Multiplication Trick #3 — How to Multiply by 25 FAST!

Here’s the third in my series of multiplication tricks. The first was a trick for multiplying by 5. The second a trick for multiplying by 15, and now this one, a trick for multiplying by 25. Anyone see a pattern?


WHAT THE TRICK LETS YOU DO: Quickly multiply numbers by 25.

HOW YOU DO IT:  The key to multiplying by 25 is to think about quarters, as in “nickels, dimes, and quarters.”

Since four quarters make a dollar, and a dollar is worth 100 cents, the concept of quarters helps children see that 4 x 25 = 100.

Since four quarters make one dollar, children can see that twice that many quarters, 8, must make two dollars (200 cents). And from that fact children can see that 8 x 25 = 200.

Following this pattern, children can see that twelve quarters make three dollars (300 cents). So 12 x 25 = 300. And so on.

Fine. But how does all of this lead to a multiplication trick?

The trick is this. To multiply a number by 25, divide the number by 4 and then tack two 0s at the end, which is the same as multiplying by 100.

A few more examples:

16 x 25. Divide 16 by 4 to get 4, so the answer is 400. [In money terms, 16 quarters make $4 = 400 cents.]

24 x 25. Divide 24 by 4 to get 6, so the answer is 600. [In money terms, 24 quarters make $6 = 600 cents.]

48 x 25. Divide 48 by 4  to get 12, so the answer is 1200. [In money terms, 48 quarters make $12 = 1200 cents.]

Try these for practice:

20 x 25

32 x 25

36 x 25

16 x 25

24 x 25

44 x 25

52 x 25

76 x 25


 20 x 25  =  500

32 x 25  =  800

36 x 25  =  900

16 x 25  =  400

24 x 25  =  600

44 x 25  =  1100

52 x 25  =  1300

76 x 25  =  1900

But wait, you protest … what about all of the numbers that are not divisible by 4? Good question! But it turns out that there’s a workaround. You still divide by 4, but now you pay attention to the remainder.

If the remainder is 1, that’s like having 1 extra quarter, an additional 25 cents, so you add 25 to the answer.

Example:  17 x 25. Since 17 ÷ 4 = 4 remainder 1, the answer is 400 + 25 = 425.

If the remainder is 2, that’s like having 2 extra quarters, an additional 50 cents, so you add 50 to the answer.

Example: 26 x 25. Since 26 ÷ 4 = 6 remainder 2, the answer is 600 + 50 = 650.

If the remainder is 3, that’s like having 3 extra quarters, an additional 75 cents, so you add 75 to the answer.

Example:  51 x 25. Since 51 ÷ 4 = 12 remainder 3, the answer is 1200 + 75 = 1275.

Now try these for practice:

9 x 25

11 x 25

14 x 25

19 x 25

22 x 25

25 x 25

34 x 25

49 x 25


9 x 25  =  225

11 x 25  =  275

14 x 25  =  350

19 x 25  =  475

22 x 25  =  550

25 x 25  =  625

34 x 25  =  850

49 x 25  =  1225

Happy teaching!

—  Josh

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  Just click the links in the sidebar for more information! 

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Subtracting Integers

A reader named Michelle said she enjoyed my post on Memorizing the Times Tables. And then she asked if I have any tips on teaching students to SUBTRACT INTEGERS.

It turns out that the answer is “yes,” and there are two places where I model this topic.

The first is an excerpt from my Algebra Survival Guide, an excerpt about subtracting integers that you can check out now.

First click on this link, then scroll down to read pages 43-46 (you can enlarge the print size by increasing the percentage located in the top bar):

But there’s more:

I wrote an entire book on the topic of combining integers, PreAlgebra Blastoff! We created foam manipulatives to get across the idea of integers. There’s a piece of a foam with a hole in the center, the NEGATon, which stands for – 1. There’s a piece that fills the hole, the POSITon, and that manipulative stands for + 1. When you put the POSITon inside the NEGATon, you get 0, and that is a piece called a ZERObi.

Using these manipuatives you can model and teach students how to combine integers, and add and subtract integers.

To learn more about this system and to see how the manipulatives work, go to this website:

Let me know if you have any questions on this topic. It’s certainly an important issue.

Happy teaching!

— Josh

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Hopes for Obama and Education

Here’s another interactive post for all of you blog readers.

It’s sort of the Pink Elephant that has landed in the living room, and I can’t just pretend it’s not here any more

The election of Barack Obama could portend significant changes in the system of public education in this country.

If you’re open to sharing, I’d really like to hear your thoughts.

What do you hope for in an Obama Administration, with regard to education? Research and new Grants? Changes to No Child Left Behind? Greater funding for urban and rural districts? Higher teacher pay? An end to vouchers? More accountability? Merit pay?

What are your concerns?

Do you have any ideas that you think Obama would do well to heed?

What might Obama ignore in the field of education that he would do well to pay close attention to?

I’m opening this up pretty wide. But with one restriction. All comments must be on the topic of education. Any that are not on education I will have to discard.

Your coments may be short, Just, what’s on the top of your your mind? I’d really like to know. And we can all benefit by hearing from one anoher.


— Josh

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Fraction divided by a fraction


This one is going out to all of you … on station MCHT.

Does anyone know any good tricks for the situation where a fraction divides another fraction. In other words, for a problem like [(2/3)÷(4/5}, does anyone know of a col way to make this much easer than the way most people learn this in a school?

If so, send me your thoughts In any case, after I get a bunch of your ideas, I’ll share mine. Then we can vote on which approach we like the most.

If you’ve got an idea, send it to:

Make the subject line: Dividing a Fraction by a Fraction.

Have fun!

— Josh

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The problem:

It’s your friend’s 73rd birthday. You’ve put together a surprise party and baked a special coconut meringue cake. But at the last minute you realize that — golly gee! — you forgot to get candles.

Rummaging through your drawers with just five minutes before your friend’s scheduled arrival, you find that you do have 14 candles. And being a brilliant mathematician, you realize that you can represent the number 73 with these 14 candles, using every candle. How do you do it?

As a hint, here’s a model showing how to do a problem like this, if you are celebrating someone’s 44th birthday, when you have just 13 candles. Notice that each dot on the top row is one candle.


Note that you may use icing to create the symbols: +, –, x, ÷, and you may also put in exponents, using candles to show the value of the exponent.

Have fun!

Send answers to:

Make the Subject line: POTM

Please include your full name, where you live, and if you don’t mind, describe your connection to math and math education (for example: teacher, tutor, math enthusiast, etc.).

The first person to send in a correct answer receives a $20 gift certificate toward the purchase of any Singing Turtle Press products. I’ll fill the winner in on the details by email.

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ANSWER: October Problem of the Month

The problem, once again, was this:

Here’s the OCTOBER POTM —


A) Each of the variables x, y, and z, stands for a unique natural number from 1 thru 9 inclusive.

B) The following, therefore, represents a number in the millions place: 3,x2y,1z3

C) This number: 3,x2y,1z3 is divisible by 9.

Question: What are three other 7-digit numbers containing all of these same digits: 3, x, 2, y, 1, z, and 3 that are also divisible by 9?

The correct answer was provided by Kevin Pickard, 48, a computer programmer and homeschooling parent of “two very smart girls,” ages 7 and 10. Kevin lives in Markham, ON, Canada.
Here is Kevin’s answer, in his own words:
Given that 3,x2y,1z3 is divisible by 9, then all of its digits
will add up to 9 eventually by repeated sums (eg. if x=4, y=2, z=3
then 3 + 4 + 2 + 2 + 1 + 3 + 3 = 18 and 1 + 8 = 9). So any arrangement
of these digits will still add up to 9 by repeated sums. These new
arrangements will therefore also be divisible by 9. So 3 other 7-digit
numbers that are also divisible by 9 would be as follows.
Plugging in the example values of x,y & z from above confirms
the result.
Josh’s note: In this problem, ANY new combination of the 7 given digits will produce a number divisible by 9. So in a sense this is a “trick problem.” The way the problem was written made some people think that only certain combinations of the digits would result in a number divisible by 9. The problem shows that the rule for divisibility by 9 is very flexible. When applying this divisibility rule, you need not take into account the order of the digits whatsoever. All that matters is that the sum of the digits is divisible by 9.

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