See also the similar post by Ma and Pa Kettle.

*Posted by John Quintanilla on September 5, 2016*

https://meangreenmath.com/2016/09/05/abbott-costello-and-division/

I normally hate these math-is-hard jokes, but this one made me laugh. Forgiving somebody is harder than computing ?

I don’t know who created this cartoon; I’ll be happy to credit it if anybody knows.

*Posted by John Quintanilla on February 9, 2016*

https://meangreenmath.com/2016/02/09/seventy-times-seven/

I absolutely love this. There’s more to teaching multiplication than rote memorization, but this is wonderful reinforcement for what happens inside of a classroom.

And also this:

*Posted by John Quintanilla on January 30, 2016*

https://meangreenmath.com/2016/01/30/teaching-multiplication/

Last March, on Pi Day (March 14, 2015), I put together a mathematical magic show for the Pi Day festivities at our local library, compiling various tricks that I teach to our future secondary teachers. I was expecting an audience of junior-high and high school students but ended up with an audience of elementary school students (and their parents). Still, I thought that this might be of general interest, and so I’ll present these tricks as well as the explanations for these tricks in this series. From start to finish, this mathematical magic show took me about 50-55 minutes to complete. None of the tricks in this routine are original to me; I learned each of these tricks from somebody else.

Though this wasn’t part of my Pi Day magic show, I recently read an interesting variant on my fourth trick. The next time I do a mathematical show, I’ll do this trick next — not to amaze and stun my audience, but to see if my audience can figure out why it works. Each member of the audience will need to have a calculator (a basic four-function calculator will suffice). Here’s the patter:

I want you to take out your calculator. Using only the digits 1 through 9, there are three rows, three columns, and two diagonals. I want you to pick either a row, a column, or a diagonal. Then I want you to enter a three digit number using those numbers. For example, if you chose the first row, you can enter 123 or 312 or 231 or any three-digit number using each digit once.

Now, I want you to multiply this number by another three-digit number. So hit the times button.

(pause)

Now, choose another row, column, or diagonal and type in another three-digit number, using each of the three digits once.

(pause)

Now hit the equals button to multiply those two numbers together.

(pause)

Is everyone done? The product you just computed should have either five or six digits. I want you to concentrate on one of those digits. Just make sure that you concentrate on a digit other than zero, because zero is boring. So concentrate on a nonzero digit.

(pause)

(I point to someone.) Without telling me the digit you chose, please tell me the other digits in your product.

The audience member will say something like, “3, 7, 9, and 2.” To which I’ll reply in three seconds or less, “The number you chose was 6.”

Then I’ll turn to someone else and ask which numbers were not scratched out. She’ll say something like, “1, 1, 9, 7, and 2.” I’ll answer, “The number you chose was 7.”

And then I’ll repeat this a few times, and everyone’s amazed that I knew the different numbers that were chosen.

Clearly this works using the same logic as my fourth magic trick: the product is always a multiple of 9, and so I can add the digits to figure out the missing digit. The more interesting question is: Why is the product always a multiple of 9?

This works because each of the factors of the product is a multiple of 3. Let’s take another look at the calculator.

If the first row is chosen, the sum of the digits is 1+2+3 = 6, a multiple of 3. And it doesn’t matter if the number is 123 or 312 or 231… the order of the digits is unimportant.

If the second row is chosen, the sum of the digits is 4+5+6 = 15, a multiple of 3.

If the third row is chosen, the sum of the digits is 7+8+9 = 24, a multiple of 3.

If the first column is chosen the sum of the digits is 1+4+7=12, a multiple of 3.

If the second column is chosen, the sum of the digits is 2+5+8 = 15, a multiple of 3.

If the third column is chosen, the sum of the digits is 3+6+9 = 18, a multiple of 3.

If one diagonal is chosen, the sum of the digits is 1+5+9 = 15, a multiple of 3.

If the other diagonal is chosen, the sum of the digits is 3+5+7 = 15, a multiple of 3.

This can be stated more succinctly using algebra. The digits in each row, column, and diagonal form an arithmetic sequence. For each row, the common difference is 1. For each column, the common difference is 3. And for a diagonal, the common difference is either 2 or 4. If I let be the first term in the sequence and let be the common difference, then the three digits are , , and , and their sum is

,

which is a multiple of 3. (Indeed, the sum is 3 times the middle number.)

So each factor is a multiple of 3. That means the product has to be a multiple of . In other words, if the first factor is and the second factor is , where and are integers, their product is equal to

,

which is clearly a multiple of 9. Therefore, I can use the same adding-the-digits trick to identify the missing digit.

*Posted by John Quintanilla on November 15, 2015*

https://meangreenmath.com/2015/11/15/my-mathematical-magic-show-part-5d/

Last March, on Pi Day (March 14, 2015), I put together a mathematical magic show for the Pi Day festivities at our local library, compiling various tricks that I teach to our future secondary teachers. I was expecting an audience of junior-high and high school students but ended up with an audience of elementary school students (and their parents). Still, I thought that this might be of general interest, and so I’ll present these tricks as well as the explanations for these tricks in this series. From start to finish, this mathematical magic show took me about 50-55 minutes to complete. None of the tricks in this routine are original to me; I learned each of these tricks from somebody else.

Though this wasn’t part of my Pi Day magic show, I recently read an interesting variant on my fourth trick. The next time I do a mathematical show, I’ll do this trick next — not to amaze and stun my audience, but to see if my audience can figure out why it works. Each member of the audience will need to have a calculator (a basic four-function calculator will suffice). Here’s the patter:

I want you to take out your calculator. Using only the digits 1 through 9, there are three rows, three columns, and two diagonals. I want you to pick either a row, a column, or a diagonal. Then I want you to enter a three digit number using those numbers. For example, if you chose the first row, you can enter 123 or 312 or 231 or any three-digit number using each digit once.

Now, I want you to multiply this number by another three-digit number. So hit the times button.

(pause)

Now, choose another row, column, or diagonal and type in another three-digit number, using each of the three digits once.

(pause)

Now hit the equals button to multiply those two numbers together.

(pause)

Is everyone done? The product you just computed should have either five or six digits. I want you to concentrate on one of those digits. Just make sure that you concentrate on a digit other than zero, because zero is boring. So concentrate on a nonzero digit.

(pause)

(I point to someone.) Without telling me the digit you chose, please tell me the other digits in your product.

The audience member will say something like, “3, 7, 9, and 2.” To which I’ll reply in three seconds or less, “The number you chose was 6.”

Then I’ll turn to someone else and ask which numbers were not scratched out. She’ll say something like, “1, 1, 9, 7, and 2.” I’ll answer, “The number you chose was 7.”

And then I’ll repeat this a few times, and everyone’s amazed that I knew the different numbers that were chosen.

Clearly this works using the same logic as yesterday’s post: the product is always a multiple of 9, and so I can add the digits to figure out the missing digit. The more interesting question is: Why is the product always a multiple of 9? I’ll address this in tomorrow’s post.

*Posted by John Quintanilla on November 14, 2015*

https://meangreenmath.com/2015/11/14/my-mathematical-magic-show-part-5c/

One of my guilty pleasures in the 1990s, when I was far too old to be watching children’s cartoons, was the fantastic show “Animaniacs.” In the clip below, Yakko Warner describes how to multiply 47 and 83.

*Posted by John Quintanilla on September 1, 2015*

https://meangreenmath.com/2015/09/01/multiplication-and-animaniacs/

I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on doing basic arithmetic with very large numbers that exceed the character displays of most calculators.

Part 1: Addition

Part 2: Multiplication

Part 3: Division

*Posted by John Quintanilla on February 23, 2015*

https://meangreenmath.com/2015/02/23/arithmetic-with-big-numbers-index/

I found the following blogpost very intriguing. This sounds like an engaging way of teaching children how to multiply, whether or not the Common Core was the educational fashion of the day: http://mathcoachscorner.blogspot.com/2014/08/a-peek-inside-theres-nothing-alien.html.

*Posted by John Quintanilla on January 6, 2015*

https://meangreenmath.com/2015/01/06/math-coachs-corner/

From Bleacher Report:

Houston Texans quarterback Ryan Fitzpatrick… threw for 358 yards and six touchdowns in a 45-21 victory over the Tennessee Titans on Sunday [November 30, 2014]. However, [his son] Brady was the star of the postgame press conference.

Fitzpatrick put his son on the spot at the end of the press conference. In a matter of seconds, Brady was able to multiply 93 by 97 in his head.

After the thought bubble, I’ll reveal the likely way that young Brady did this.

Here’s a trick for multiplying two numbers in their 90s which is accessible to bright elementary-school students. We begin by multiplying out :

For , we have and . So , and . So the first two digits of the product is .

Also, . So the last two digits are .

Put them together, and we get the product $100 \times 90 + 21 = 9021$.

I don’t expect that young Brady knew all of this algebra, but I expect that he did the above mental arithmetic to put together the product. Well done, young man.

*Posted by John Quintanilla on December 5, 2014*

https://meangreenmath.com/2014/12/05/texans-qb-ryan-fitzpatricks-son-shows-off-math-skills-during-postgame-press-conference-part-2/

From Bleacher Report:

Houston Texans quarterback Ryan Fitzpatrick… threw for 358 yards and six touchdowns in a 45-21 victory over the Tennessee Titans on Sunday [November 30, 2014]. However, [his son] Brady was the star of the postgame press conference.

Fitzpatrick put his son on the spot at the end of the press conference. In a matter of seconds, Brady was able to multiply 93 by 97 in his head.

I’ll reveal the (likely) way that young Brady Fitzpatrick pulled this off tomorrow. In the meantime, I’ll leave a thought bubble if you’d like to try to figure it out on your own.

*Posted by John Quintanilla on December 3, 2014*

https://meangreenmath.com/2014/12/03/texans-qb-ryan-fitzpatricks-son-shows-off-math-skills-during-postgame-press-conference-part-1/