Another poorly written word problem (Part 4)

Textbooks have included the occasional awful problem ever since Pebbles Flintstone and Bamm-Bamm Rubble chiseled their homework on slate tablets while attending Bedrock Elementary. But even with the understanding that there have been children have been doing awful homework problems since the dawn of time (and long before the advent of the Common Core), this one is a doozy.

There’s no sense having a debate about standards for elementary mathematics if textbook publishers can’t write reasonable homework problems.

My reaction to this problem is pretty much echoed by the following post: http://www.patheos.com/blogs/friendlyatheist/2015/10/22/sometimes-estimating-is-better-than-getting-the-exact-answer/:

Yes, this is an awful word problem. This should never have appeared in a math textbook or workbook. But if it appeared in a workbook, then it should never have been assigned by a teacher. And if it accidentally got assigned by a teacher, then the teacher should have extended some grace in the grading of the problem.
Even with all that said, estimation has been in the elementary curriculum for decades and is not an invention of the Common Core. Furthermore, estimation is an important skill for students to acquire. From the above website:

Suppose you’re buying groceries. You have four items in your cart that cost $1.99, $4.93, $6.03, and $5.14.

If all you have is $20 in your wallet, is that enough to pay for the items?

I think that’s a very realistic question.

It would take you at least a little bit of time to add up those numbers individually and get an exact number. Would it answer your question? Absolutely. But you don’t need an exact answer.

The smarter thing to do would be to simply round the numbers. We should be saying to ourselves, “2 + 5 + 6 + 5 equals 18… throw in some tax… and I should still be under $20.”

Why is that better? Because the exact amount doesn’t really make a difference. You just need to be close enough.

I have deep and profound theological differences with the author of this post. But on this math issue, he’s right on the money (pardon the pun).

Teaching multiplication

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:

Source: https://www.facebook.com/RealDLHughley/photos/a.247216605320684.53846.219856014723410/1096551810387155/?type=3&theater

Lessons from teaching gifted elementary school students: Index (updated)

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 various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students. (This is updated from my previous index.)

Part 1: A surprising pattern in some consecutive perfect squares.

Part 2: Calculating 2 to a very large exponent.

Part 3a: Calculating 2 to an even larger exponent.

Part 3b: An analysis of just how large this number actually is.

Part 4a: The chance of winning at BINGO in only four turns.

Part 4b: Pedagogical thoughts on one step of the calculation.

Part 4c: A complicated follow-up question.

Part 5a: Exponentiation is multiplication as multiplication is to addition. So, multiplication is to addition as addition is to what? (I offered the answer of incrementation, but it was rejected: addition requires two inputs, while incrementation only requires one.)

Part 5b: Why there is no binary operation that completes the above analogy.

Part 5c: Knuth’s up-arrow notation for writing very big numbers.

Part 5d: Graham’s number, reputed to be the largest number ever to appear in a mathematical proof.

Another poorly written word problem (Part 3)

There’s no sense having a debate about standards for elementary mathematics if textbook publishers can’t construct sentences that can be understood by elementary students.

My Mathematical Magic Show: Part 7

This mathematical trick, which may well be the best mathematical magic trick ever devised, was not part of my Pi Day magic show. However, it should have been. Here’s a description of the trick, modified from the description at http://mathoverflow.net/questions/20667/generalization-of-finch-cheneys-5-card-trick:

The magician walks out of the room. A volunteer from the crowd chooses any five cards at random from a deck, and hands them to your assistant so that nobody else can see them. The assistant glances at them briefly and hands one card back, which the volunteer then places face down on the table to one side. The assistant quickly place the remaining four cards face up on the table, in a row from left to right. After all of this is completed, the magician re-enters the room, inspects the faces of the four cards, and promptly names the hidden fifth card.

In turns out that the trick is a clever application of permutations (there are $3! = 6$ possible ways of ordering 3 objects) and the pigeon-hole principle (if each object belongs to one of four categories and there are five objects, then at least two objects must belong to the same category). These principles from discrete mathematics (specifically, combinatorics) make possible the Fitch-Cheney 5-Card Trick.

Unlike the other tricks in this series, the Fitch-Cheney 5-Card Trick requires a well-trained assistant (or a smartphone app that plays the role of the assistant).

A great description of how this trick works can be found at Math With Bad Drawings. For a deeper look at some of the mathematics behind this trick, I give the following references:

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 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 $a$ be the first term in the sequence and let $d$ be the common difference, then the three digits are $a$ , $a + d$ , and $a + 2d$ , and their sum is

$a + (a+d) + (a+ 2d) = 3a + 3d = 3(a+d)$ ,

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 $9$ . In other words, if the first factor is $3m$ and the second factor is $3n$ , where $m$ and $n$ are integers, their product is equal to

$(3m)(3n) = 9(mn)$ ,

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

My Mathematical Magic Show: Part 5c

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.

My Mathematical Magic Show: Part 5b

Here’s the patter for my fourth and most impressive trick. As before, my audience has a sheet of paper and a pen or pencil; quite a few of them have calculators.

Write down any five-digit number you want. Just make sure that the same digit repeated (not something like 88,888).

(pause)

Now scramble the digits of your number, and write down the new number. Just be sure that any repeated digits appear the same number of times. (For example, if your first number was 14,232, your second number could be 24,231 or 13,422.)

(pause)

Is everyone done? Now subtract the smaller of the two numbers from the bigger, and write down the difference. Use a calculator if you wish.

(pause)

Has everyone written down the difference. Good. Now, pick any nonzero digit in the difference, and scratch it out.

(pause)

(I point to someone.) Which numbers did you not scratch out?

The audience member will say something like, “8, 2, 9, and 6.” To which I’ll reply in three seconds or less, “The number you scratched out was a 2.”

Then I’ll turn to someone else and ask which numbers were not scratched out. She’ll say something like, “3, 2, 0, and 7.” I’ll answer, “You scratched out a 6.”

After performing this trick, I’ll explain how it works. I gave a very mathematical explanation in a previous post for why this trick works, but the following explanation seems to go over well with even elementary-school students. I’ll ask an audience member for the two five-digit numbers that they subtracted. Suppose that she tells me that hers were

$43,125-24,513$

I’ll now tell the audience that, ordinarily, we would plug this into a calculator or else start by subtracting the ones digits. However, I tell the audience, I’m now going to write this in a very unusual way:

$(40,000 + 3,000 + 100 + 20 + 5) - (20,000 + 4,000 + 500 + 10 + 3)$

I tell the audience, “For now, I’m not saying why I did this. But does everyone agree that I can do this?” Once I get agreement, then I proceed to the next step by grouping like digits together:

$(40,000 - 4,000) + (3,000 - 3) + (100 - 10) + (20 - 20,000) + (5 - 500)$

Again, I tell the audience, “For now, I’m not saying why I did this. But does everyone agree that I can do this?” Once I get agreement, then I proceed to the next step by reversing the signs of any negative differences:

$(40,000 - 4,000) + (3,000 - 3) + (100 - 10) - (20,000 - 20) - (500 - 5)$

Next, I factor each common difference. Notice that in each parenthesis, the second number is a factor of the first number:

$4,000(10-1) + 3(1,000 - 1) + 10(10 - 1) - 20(1,000 - 1) - 5(100 - 1)$ ,

$4,000(9) + 3(999) + 10(9) - 20(999) - 5(99)$ .

Notice that the number in each pair of parentheses is a multiple of 9. Therefore, no matter what, the difference must be a multiple of 9.

This is the key observation that makes the trick work. Now, I go back to my audience member and ask what the difference actually was:

$43,125-24,513 = 18,612$

This difference must be a multiple of 9. Therefore, by one of the standard divisibility tricks, the digits of this number must add to a multiple of 9:

$1 + 8 + 6 + 1 + 2 = 18$ .

Then I’ll ask the audience member, “Which number did you scratch out?” Suppose she answers 6. Then I’ll add up the remaining numbers:

$1 + 8 + 1 + 2 = 12$ .

So I ask the audience, “So these four numbers add up to 12, but I know that all five numbers have to add up to a multiple of 9. What’s the next multiple of 9 after 12?” They’ll answer, “18”. I ask, “So what does the missing number have to be?” They’ll answer “18-12, or 6.”

Then I’ll repeat with someone else. If an audience member answers “8, 2, 9, and 6,” I’ll ask the audience for the sum of these four numbers. (It’s 25.) So they can figure out that the scratched-out number was 2, since 25+2 = 27 is the next multiple of 9 after 25.

I’m often asked why I made people choose a five-digit number at the start of the routine. The answer is, I could have chosen any size number I wanted as long as I’m comfortable with quickly adding the digits at the end of the magic trick. In other words, if I had permitted nine-digit numbers, I might need to add 8 numbers at the end of the routine to get the missing number. I could do it, but I wouldn’t get the answer as quickly as the five-digit numbers.

Also, I’m often asked why it was important that I told the audience to scratch out a nonzero number. Well, suppose that I came to end of the routine and the audience member told me her remaining digits were 4, 3, and 2. These numbers have a sum of 9, and so the missing number hypothetically could be 0 or 9. So by instructing the audience to not scratch out a 0, that eliminates the ambiguity from this special case.

After showing the audience how the trick works, I’ll then ask an audience member to come forward and repeat the trick that I just performed. Then I’ll move on to the final act of my routine, which I’ll present in tomorrow’s post.

My Mathematical Magic Show: Part 5a

Here’s the patter for my fourth and most impressive trick. As before, my audience has a sheet of paper and a pen or pencil; quite a few of them have calculators.

Write down any five-digit number you want. Just make sure that the same digit repeated (not something like 88,888).

(pause)

Now scramble the digits of your number, and write down the new number. Just be sure that any repeated digits appear the same number of times. (For example, if your first number was 14,232, your second number could be 24,231 or 13,422.)

(pause)

Is everyone done? Now subtract the smaller of the two numbers from the bigger, and write down the difference. Use a calculator if you wish.

(pause)

Has everyone written down the difference. Good. Now, pick any nonzero digit in the difference, and scratch it out.

(pause)

(I point to someone.) Which numbers did you not scratch out?

The audience member will say something like, “8, 2, 9, and 6.” To which I’ll reply in three seconds or less, “The number you scratched out was a 2.”

Then I’ll turn to someone else and ask which numbers were not scratched out. She’ll say something like, “3, 2, 0, and 7.” I’ll answer, “You scratched out a 6.”

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

Then I explain how this trick works, which I’ll do in tomorrow’s post.

My Mathematical Magic Show: Part 4d

For my third trick, I’ll present something that I first saw when pulling Christmas crackers with my family. I’ll give everyone a piece of paper with six cards printed. I’ll also have a large version of this paper shown at the front of the room (taken from http://diaryofagrumpyteacher.blogspot.com/2014/04/freebie-friday-magic-number-cards.html; see also this Google search if this link somehow goes down):

Here’s the patter:

Think of a number from 0 to 63. Then, on your piece of paper, circle the cards that contain your number. For example, if your number is 15, you’ll need to circle the card in the upper-left because 15 is on that card. You’d have to circle all the cards that contain 15.

(pause)

Is everyone done? (Points to someone) Which cards did you circle?

At this point, the audience member will say something like “Top left, top middle, and bottom right.” Then I will add the smallest numbers on each card (in this case, 1, 2, and 32) and answer in five seconds or less, “Your number was 35 (or whatever the sum is).” It turns out that the number is always the sum of the smallest numbers on the selected cards.As shown in yesterday’s post, this is a consequence of the binary representation of whole numbers (as opposed to the ordinary decimal representation).

Though I don’t do this in my magic routine for the sake of time, I have challenged my future high school math teachers to develop a similar magic trick for some other base, like base 3, just to make sure that they really understand the concept behind the above magic trick. Here are the cards that work for base 3 (taken from http://www.mathman.biz/html/sherimagic.html).

I encourage the reader to develop another set of cards for base 5. It will require 10 cards for numbers from 1 to 24.

With tomorrow’s post, I’ll continue my description of my magic routine.