Monday, August 19, 2013

Math Study

Summer vacation is waning fast, and it's time to get back to school.

For my first installment of the new school year, I'd like to talk a little bit about an aspect of test preparation that many people do not take seriously enough:  Sleep.

I know that some students take a very last-minute approach to preparing for tests (or finishing projects, or any number of other school-related activities).  I have been in this position myself, and it is quite stressful.

As {insert name of deadline here} approaches, people discover that they can't fit all of the things they should be doing into the amount of time that is available to them.  All too often, the thing that gets discarded is sleep.

I'm not going to go into the health issues associated with sleep deprivation... needless to say, you'll be healthier if you get enough sleep.  Also, there are the obvious safety concerns... sleep-deprived drivers can be just as dangerous as drunk drivers.

My goal here is to draw attention to the academic ramifications of missing out on sleep.

Personally, I think that students need to consider the type of academic activity they are preparing for when deciding how much sleep to cut out of their schedule.  Students need to understand that skipping sleep will affect their performance.  The goal is to find an acceptable trade-off between needed preparation and lower performance quality.

Some activities do not require much performance.  For example, if a big project is due then the student only needs to be able to stumble into the classroom and turn the project in.  Students can afford to miss more sleep in a situation like this.

However, if the student will be required to present his/her project to the class (and will be graded on the presentation) then more sleep is advised.  It doesn't make much sense to stay up all night to enhance the quality of a project if the presentation ends up being terrible.

When preparing for a test, there is typically a point at which the loss of brain power due to sleep deprivation outweighs the benefits of more cramming.  Continuing to stay awake beyond this point does more damage than good.

This gets a bit tricky because different people need different amounts of sleep.  Therefore, I can't possibly prescribe a last-minute cramming strategy that will work for everyone...

...and I guess that's the real point of this post:  If you plan ahead and don't procrastinate, then you won't have to worry about staying up and losing a significant amount of sleep the night before a big presentation or exam.

My latest Silly Math Song was inspired by this concept:

If you are a student, parent, or teacher, I hope you have a wonderful school year.

For math tutorials and silly math songs, visit

Friday, May 10, 2013

Deconstructing Chocolate Math

Every year, mass emails go out challenging people to calculate their age using “Chocolate Math”.  In a minute, I’ll deconstruct the 2013 version of this fun math activity – but first, try out this (simpler) activity:

I’m going to read your mind, with the help of a little mathematical trickery.

1.      Think of a number – any number will work, but it’s easiest if you choose a smaller number (between 1 and 10).

2.      Add 7.

3.      Take the result and add your original number.

4.      Take the result and add 9.

5.      Take the result and divide it by 2.

6.      Take the result and subtract your original number.

The number you are now thinking of is… … … 8.

This result always holds (assuming you do the math correctly), regardless of your choice of starting number.

Do you know why?  (If not, don’t worry.  I’ll give the explanation a little later.)

Now, let’s take a look at Calculating your Age by using Chocolate Math – 2013 Version.

Follow these steps to use chocolate math to calculate your age:

1.      Pick the number of times a week that you would like to have chocolate.  (More than once, but less than 10 times.)

2.      Multiply this number by 2

3.      Add 5

4.      Multiply the result by 50 (followed by some joke about getting a calculator)

5.      If you have already had your birthday this year, add 1763.  If not, add 1762.

6.      Subtract the four-digit year that you were born.

You now have a three-digit number.  The first digit is your original number (i.e. how many times you want to have chocolate each week).  The next two numbers equal… YOUR AGE!

To discover why this works (and why this version will only work in the year 2013), we’ll need to make use of some basic Algebra skills.

Every step of this activity is dictated to you, EXCEPT the first step… there is no way to know how many times you will decide you want chocolate each week.

So – let’s define a variable.  Let  x  represent the number of times you want chocolate each week.

By writing algebraic expressions for each of the following steps, we can discover what is happening.

Pick a number:                                    x

Multiply by 2                                      2x

Add 5                                                  2x + 5

Multiply by 50                                    50(2x + 5) = 100x + 250

At this point, you have a choice based on whether or not you’ve had your birthday this year.

If yes, add 1763                                  100x + 250 + 1763 = 100x + 2013

If no, add 1762                                   100x + 250 + 1762 = 100x + 2012

The expression now contains 100x plus the last year you had a birthday.

By subtracting your birth year, you will end up with 100x + your age.  Therefore, the first digit of your three-digit number will be the number of times you want chocolate each week, and the last two digits will be your age.

(By the way, this activity doesn’t work for people who are more than 99 years old… how discriminatory!!!)

Here’s an interesting thought – there are many ways to force an algebraic expression of either 100x + 2013  or  100x + 2012.  This means that, if you were so inclined, you could make up your own set of steps that would calculate someone’s age using Chocolate Math.

Now – if you understand the math behind the Chocolate Math activity, you can probably follow the steps in the initial activity to discover why those steps will always yield a result of 8.  If you don’t want to work through the process on your own, here it is:

Think of a number                               x

Add 7                                                  x + 7

Add your original number                  x + 7 + x = 2x + 7

Add 9                                                  2x + 7 + 9 = 2x + 16

Divide by 2                                         (2x + 16) ÷ 2 = x + 8

Subtract your original number            x + 8 – x = 8

With some basic algebra skills, you too can read minds!
For math tutorials and Silly Math Songs, visit:

Thursday, March 28, 2013

Factoring Song

I've got an NBA-themed article in the works, but decided to postpone finishing it in order to work on my latest Silly Math Song.

FACTORING SONG: FACTOR GANGNAM STYLE teaches the basics of factoring, as well as reducing fractions that contain polynomials.

The information is pretty rapid-fire as the song covers all of these types of factoring:

Factoring out the GCF
Factoring the Difference of Squares
Factoring Quadratic Trinomials with a Leading Coefficient of 1
Factoring by Grouping
Factoring the Sum and Difference of Cubes
Factoring General Quadratic Trinomials (ax^2 + bx + c)

It's interesting to note that the vast majority of quadratic expressions are not factorable - but factoring is such a great exercise in mathematical logic, and has been part of the core Algebra curriculum for years.

Hope you enjoy the song.


For Math Tutorials and more Silly Math Songs, visit

Thursday, February 28, 2013

The Road Not Taken (Logic Puzzle)

In Robert Frost’s famous poem, The Road Not Taken (sometimes mistakenly referred to as “The Road Less Travelled”), a traveler is faced with the difficult choice of which road he should follow.

This brings to mind a classic logic puzzle, with mathematical implications.

Here is my poetic interpretation of the puzzle.

THE ROAD NOT TAKEN – Mr. Wagneezy Version
(Line 1 by Robert Frost)

Two roads diverge in a yellow wood
One leads to certain death
The other leads to riches untold
I stop to catch my breath

I soon discover I have no clue
Exactly which road is which
I look to the right, and then to the left
My eyes begin to twitch

Suddenly two gnomes appear
From out of nearby briars
One of them is a truthful gnome
The other one is a liar

In looking at these gnomes, alas
I cannot tell the difference
Which one speaks truth?   Which one speaks lies?
I fight the urge to wince

These seemingly identical gnomes
Both know which road to take
But instead of making it clear to me
They make it nearly opaque

The gnomes agree that one of them
Will answer a single question
Once I get the answer
They will end the conversation

I still can’t tell which one speaks truth
And which one is the liar
They smirk at me, these pesky gnomes
That came out of the briars

I must determine what to say
It is a daunting task
To get the gnomes to reveal the way
What question should I ask?


Successfully getting these gnomes to show us which road to choose will take careful planning.  Since we have no way of knowing which gnome tells the truth and which gnome lies, we must be able to come up with a question that both gnomes would answer the same way.

Obviously, the direct approach (e.g. “Which road leads to untold riches?”) will not work, because the truth-teller would point to one road while the liar would point to the other road.

Therefore, we must ask an indirect question – one that incorporates both the truth and the lie.  We can accomplish this by asking one gnome to tell us which road the other gnome would point us towards.  (E.g. “If I asked the other gnome which road leads to untold riches, which road would he point to?”)

The logic here is that the truth about a lie gives the same result as a lie about the truth.

More specifically, if we happen to talk to the truth-teller, he will point to the wrong road because that is the road the liar would have pointed to.

On the other hand, if we happen to talk to the liar, he will also point to the wrong road because that is not the road the truth-teller would have pointed to.

In either case, the wrong road will be indicated, and we can choose the other road to travel on.


For movie buffs – a version of this puzzle appeared in the movie Labyrinth :


Earlier, I mentioned that this puzzle has mathematical implications.

Let’s deconstruct this puzzle into a similar mathematical question.

I hope you agree with me that telling the truth is positive, and lying is negative.

Consider mathematical operations that can be performed on two numbers.  Suppose one of the numbers is positive, and the other number is negative, but we DON’T KNOW WHICH IS WHICH.

Which operations are guaranteed to give us results with the same sign, regardless of which number is positive?

If we arbitrarily choose ±2 and ±3 for our numbers and use them to explore each operation, we get the following:


As we can see, multiplication and division are the only operations that fit the bill.

(Not coincidentally, multiplication and division have the same priority in the Order of Operations.)

Therefore, the logical argument

“The truth about a lie is equivalent to a lie about the truth”

seems to match up with the mathematical concepts

“A negative times a positive is equivalent to a positive times a negative”


“A negative divided by a positive is equivalent to a positive divided by a negative”.

By the way – multiplication and division both exhibit the desired property because

1)      Multiplying is the same thing as dividing by the reciprocal.
2)      Dividing is the same thing as multiplying by the reciprocal.
3)      A number’s sign (positive or negative) is the same as the sign of the number’s reciprocal.
May all your roads be wisely chosen.
For Math Tutorials & Silly Math Songs, visit

Friday, February 8, 2013

What’s the difference between 90% and 100%? (Traits of HIGHLY successful math students)

The differences between the students who fail in math and the students who get decent grades – C and above – are generally clear cut, and easily identifiable (unless there is a learning disability involved).  Most of these differences aren’t even math specific… failing students in any subject can generally improve their grades by improving one or more of the following areas:

1)      Get organized
2)      Do (and turn in) assignments on time
3)      Be willing/able to seek help when needed
4)      Be willing to put in the time/effort necessary to be successful
5)      Practice/Prepare for tests
6)      If the people you hang out with aren’t committed to success, hang out with different people
7)      Don’t accept defeat easily (“If at first you don’t succeed, try, try again.”)

However, I recently found myself wondering about what sets the truly excellent math students apart from the merely-great math students.  I have a number of students earning a grade of A or A- (at or above 90%).

Why do a select few of these students consistently earn grades near 100% (or above 100%, if extra credit is offered) in math class?

In analyzing my students, I have noticed a few traits that differentiate merely-great math students from highly successful math students.

1.       Merely-great math students tend to believe that knowing a lot and being good at math is the most important component of good test taking.

Highly successful math students understand that knowing a lot and being good at math is actually the second most important component of test taking.  The most important component is being able to successfully communicate your knowledge and math excellence to your teacher/professor.

There are many merely-great math students who turn in tests containing problems with ambiguity in the solutions.  Their brains might have been doing all the right steps, but their work is suspect.  Highly successful math students provide complete solutions that are clearly mathematically sound.  As a result, they tend to receive higher scores on their tests.

Laziness also plays a role in this.  Many merely-great math students seem to follow the philosophy, “When in doubt, show less work.”  Most highly successful math students follow the philosophy, “When in doubt, show more work.”

2.       Merely-great math students don’t tend to place a high value on neatness.

Highly successful math students tend to make neatness a priority.

I can’t count how many times I’ve been grading work and discovered that a student got the wrong answer due to a mistake caused by poor handwriting.  I’ve seen “4” turn into “9”, “z” turn into “2”, “7” turn into “1” and a myriad of other sloppy mistakes.  These mistakes often prove to be the difference between a 95% test and a 100% test.

3.       Merely-great math students use lectures and class time to learn what they need to know in order to succeed.

Highly successful math students also use lectures and class time to learn what they need to know, but they tend to use this as a starting point in their learning process.  They are adept at looking in textbooks for additional examples in order to enhance the material covered in a lecture.  They are also eager to explore alternative methods and they have a desire to know WHY a particular method works or doesn’t work.

4.       Merely-great math students do their best to find out what material will be covered on a test.  (Some are even quite assertive in trying to get specific review topics from their instructors.)  They then review this material diligently to make sure they can complete the test successfully.

Highly successful math students emphasize material they know will be covered on a test when studying.  However, they also attempt to review/practice all of the other concepts from a chapter.  This is helpful in (at least) two ways:

1.       Since concepts are often interrelated, highly successful students gain a deeper understanding of the “key concepts” by broadening their review.

2.       Teachers who put extra credit items on their tests will often draw from these “other” concepts.  When these extra credit test items show up, highly successful students are ready for them.

5.        Merely-great math students sometimes put too much emphasis on being right at the expense of focusing on what went wrong.

Highly successful math students attempt to learn as much as possible from their mistakes so that they can avoid making the same kind of mistakes in the future.

 I can think of a number of students who are experts at “nickel-and-diming” teachers out of extra points.  They submit work that is good (but not great) and then have to verbally defend their work and try to convince the teacher that they deserve 100% credit. They rejoice when their arguments are fruitful and they receive a small increase in score… their mission has been accomplished.  Sadly, these students tend to get in the habit of submitting less-than-stellar work and find themselves arguing with their teachers a lot.

Highly successful math students, on the other hand, are willing to admit when their work is less than ideal.  They may argue the merits of their work with their teacher in an attempt to get more points, but their chief concern is learning how to produce stellar work in the future that will be above reproach.

The above list of traits is not exhaustive – there are probably more traits of highly successful math students that I’ve missed.  (If you’d like to add to my list, leave a comment below.)

Also, the above list does not come from a scientific study.  It is simply an anecdotal summary of my own personal observations.

I hope it will help you as you strive for true excellence.


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Saturday, January 19, 2013

Are you a good doctor? (Math Puzzle - A look at probability)

Here’s an interesting math puzzle involving probability…

1 out of every 10,000 citizens of a country has a deadly virus.  A blood test can be used to determine whether or not somebody has the virus.  The test is accurate 98% of the time.  There are 1 million people in the country, and each person is tested.  A man who tested positive for the virus talks to his doctor, who informs the man that there is a 98% chance that he does, in fact, have the virus.  The doctor is not correct in making this statement… what is the actual probability that the man has the deadly virus?

To see the answer, scroll down to the bottom.  If you would like help in understanding the logic behind the solution, keep reading.

First of all, we have to understand what probability is.

We know that 1 out of every 10,000 people actually have the virus.

Therefore, out of 1 million people, 100 people have the virus.  (1,000,000 ÷ 10,000 = 100)

The probability that a citizen of the country has the virus is 100 ÷ 1,000,000 = 0.0001 = 0.01%.

This probability is independent of the blood test.

Once the blood test is administered to every citizen in the country, the question becomes a bit more difficult.  We need to determine the probability that a person who tested positive for the virus does, in fact, have the virus.

So, how many people tested positive for the virus?

98% of people who have the virus tested positive.  98% of 100 is 98.

If 100 people have the virus, then 999,900 people do not have the virus. (1,000,000 – 100 = 999,900)

98% of people who don’t have the virus tested negative.  This means that 2% of people who don’t have the virus tested positive.  2% of 999,900 is 19,998.

The total number of people who tested positive is 98 + 19,998 = 20,096.

Now we can calculate the probability that the man who tested positive does, in fact, have the virus.

20,096 people tested positive for the virus.  Of these, 98 people actually have the virus.

Therefore, the probability that the man has the virus is 98 ÷ 20,096 = 0.0049 = 0.49%.

That’s less than half of one percent!

When considering this fact, it is tempting to think that the test is practically worthless – but this is not the case.

Before the test was administered, all we knew was that 0.01% (one 100th of a percent) of the population had the virus.  Now a group has been isolated in which about half of a percent has the virus.

If this group is re-tested, the probability that a person testing positive does, in fact, have the virus will be about 24%.

If the group that tests positive a 2nd time is re-tested, the probability that a person testing positive does, in fact, have the virus will be about 92%.

It’s amazing to see how quickly accuracy is gained through the process of re-testing.

And now you know a little bit of the mathematics that goes into being a good doctor.


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Wednesday, January 9, 2013

How NOT to cut a pizza into 10 equally-sized pieces

How do you cut a pizza into 10 pieces that are the same size?  Skip to the bottom to find out.  However, if you want to know how NOT to cut a pizza into 10 equally-sized pieces – and maybe learn a little math in the process – keep reading.

In addition to teaching math, I also sponsor a high school class.  Recently, the class I sponsor held a fundraiser at which pizza was sold.

Normally, when pizzas are delivered for our fundraisers, each pizza is cut into 8 pieces that are roughly the same size.

However, the pizzas for this fundraiser had all been cut into 10 pieces.  I was somewhat annoyed – and somewhat amused – to see that the person who had cut the pizzas had done a very poor job of making equally-sized pieces.  Some pieces appeared to be oversized, while others were quite obviously undersized.

Naturally, everybody wanted to buy the oversized pieces, while nobody wanted to buy the undersized pieces.

I began to think about the skills that a person needs to have in order to be a good pizza cutter:

            1.  Good hand-eye coordination
            2.  Good motor skills
            3.  Good math skills

…Huh?  Good math skills?!

That’s right.  The difference between mediocrity and greatness in the field of pizza cutting boils down to math.

In order to be a mediocre pizza cutter, one only needs to be able to cut a pizza (or pieces of pizza) in half.  Most people can do this well enough.  Unfortunately, if this is the extent of one’s pizza cutting skills, then it will only be possible to cut pizzas into a number of equally-sized pieces that is a power of 2.

Powers of 2 are:  21 = 2,  22 = 4,  23 = 8,  24 = 16,  etc…

If you are considering how much of the pizza each piece makes up, it’s better to consider powers of  one-half.

Since 8 is a power of two, even a mediocre pizza cutter can cut a pizza into 8 pieces that are roughly the same size.

On the other hand, in order to cut a pizza into 10 equally-sized pieces one must know that

{If you hate fractions, consider the fact that the prime factorization of 10 is 2·5.}

Therefore, after the initial cut dividing the pizza into halves, the pizza cutter must divide each half into five equally-sized pieces.  If the person cutting the pizza makes the mistake of cutting each half into two equally-sized pieces, then there is no chance of ending up with 10 equally-sized pieces because five is not divisible by two.  (This is how NOT to cut a pizza into 10 equally-sized pieces.)

This appears to be where our pizza cutter made his/her fatal mistake.  (Mathematically fatal, that is.)

A good, mathematically sound pizza cutter would have made the first cut dividing the pizza into halves.  Then, he/she would have begun cutting the halves into fifths.

If you really want to delve into the mathematical implications of this series of cuts, consider how each successive cut takes off a bigger fraction of the big piece of pizza remaining:

One last thing to consider:  I believe that people who cut pizzas for the big pizza chains (Domino’s, Pizza Hut, etc.) use a cutting instrument that reaches all of the way across a pizza – a diameter, assuming that the cut passes through the center.  If this is the case, then it is impossible to cut a pizza into an odd number of pieces by making cuts through the center.

If you’re feeling devious, try ordering a pizza and requesting that it be cut into nine pieces.

On the other hand, maybe you shouldn’t… the person cutting the pizza might actually try to do it.

***Disclaimer:  I love pizza.  I think that pizza should be one of the main food groups.  Therefore, by extension, I love the people who prepare pizza for me – even if they don’t cut the pizza into equally-sized pieces.  If you are a professional pizza cutter, I hope that you are not offended by my mathematical musings.  On the contrary, I salute you and thank you for all that you do for pizza lovers like me.***

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Thursday, January 3, 2013

Is 2013 a prime year? (A look at divisibility)

Happy New Year!  I hope your year is off to a great start.

If you like to contemplate numbers like I do, you may have wondered whether or not 2013 is a prime number. After a quick review of basic divisibility rules (yeah, there's a song for that) we can see that 2013 is divisible by 3.

(2 + 0 + 1 + 3 = 6, and 3 goes into 6.)

Factoring out 3, we have:  2013 = 3(671)

So, is 671 a prime number?  It doesn’t fit any of the divisibility rules for 2 through 10.  However, as it turns out, 671 is divisible by 11.

Factoring out 11, we have:  671 = 11(61)

Therefore, the prime factorization of 2013 is 3·11·61.

In general, how can we tell whether or not a number is divisible by 11?

Here's a fairly easy Divisibility Rule for 11:

1) Find the sum of every other digit in the number (1st digit + 3rd digit + 5th digit + …)

2) Find the sum of the digits not used in step 1 (2nd digit + 4th digit + 6th digit + …)

If these sums differ by 0, or a multiple of 11, then the original number is divisible by 11.

{If you want to know why this rule works, continue reading.  If you don't care why it works, but would like to see another neat divisibility rule for 11, skip to the bottom.}
Why this rule Works

Think back to elementary school, when you used pencil and paper to multiply.  What does it look like when a number is multiplied by 11?

Here is an example:                 12345
                                           ×         11
                                          +  123450


Let’s take a look at a general case, in which a 5-digit number is multiplied by 11.  The conclusions we draw can be generalized for multiplying any length of number by 11.

                                             ×           11
                                         +  ABCDE0


The answer will have the following values for its digits:

            1st digit:           A
            2nd digit:          (A + B)
            3rd digit:           (B + C)
            4th digit:           (C + D)
            5th digit:           (D + E)
            6th digit:           E
The sum of digits 1, 3 and 5 is A + (B + C) + (D + E)

The sum of digits 2, 5, and 6 is (A + B) + (C + D) + E

Clearly these sums have a difference of 0.

But… what if we need to carry when adding ABCDE + ABCDE0?

Exactly what happens when we carry?

If a digit’s value exceeds 9, then we subtract 10 from that value, and add 1 to the value of the digit to the left.

For example:               39
                                  +  7

 (9 + 7 = 16.  16 – 10 = 6.   We write 6 in the one’s place, and add 1 to the digit in the ten’s place)

Subtracting 10 from a digit’s value and adding 1 to the value of the digit to the left causes a net change of 11 between the sum of digits 1, 3 & 5 and the sum of digits 2, 4 & 6.

Let’s assume that in our generalized example, (D + E) exceeds 9.  In this case, since we have to carry, the answer will have the following values for its digits:

            1st digit:           A
            2nd digit:          (A + B)
            3rd digit:           (B + C)
            4th digit:           (C + D) + 1
            5th digit:           (D + E) – 10
            6th digit:           E
Now the sum of digits 1, 3 and 5 is  A + B + C + D + E – 10

And the sum of digits 2, 4 and 6 is  A + B + C + D + E + 1

These sums differ by 11.

Every time we carry (subtract 10 from a digit’s value and add 1 to the value of the digit to the left) we are causing a net change of 11 between the sum of digits 1, 3 & 5 and the sum of digits 2, 4 & 6.  These changes may accumulate, or cancel each other out.

If we carry twice, the sums could differ by 22 (if the net changes of 11 accumulate) or they could differ by 0 (if the net changes cancel each other out).

By generalizing this explanation, we can show that for any multiple of 11, the sum of digits 1, 3, 5, etc. will differ from the sum of digits 2, 4, 6, etc. by 0, or a multiple of 11.

Here’s an even easier divisibility rule for 11:

Subtract the last digit from the number formed by the remaining digits.  (You can repeat this step as often as necessary.)  If the resulting number is divisible by 11, then the original number is divisible by 11.

(Example:  671 is divisible by 11 because 67 – 1 = 66, and 66 is divisible by 11.)

Why does this work?  I bet you can figure it out…

*Hint: It might be helpful to once again consider this problem:
                                             ×           11
                                         +  ABCDE0

If you want to try out an explanation, feel free to post a comment below.

Here’s to a great (not to mention, mathtastic) year in 2013.


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