Last week, Discovery Education and the National Basketball Association (NBA) announced a partnership in which real-time data from stats.nba.com will flow into Math Techbook, and students will use that data to solve problems.
How cool is that?
Eighty students from John Hayden Johnson Middle School in Washington, DC, participated in the event, which was emceed by Hall-of-Famer “Big” Bob Lanier and made silly by Washington Wizards mascot G-Wiz.
The event received a lot of press coverage, and as you may have heard, there’s no such thing as bad publicity. But one of the articles quoted me as saying:
It’s not like a beautiful, traditional math problem.
That is not what I said. I am absolutely certain that I have never used the words beautiful and traditional in the same sentence. Well, perhaps when referring to a wedding dress or an Irish cottage, maybe, but certainly not when referring to a math problem.
I was also quoted as saying:
It’s going to be messy, for sure.
That is, in fact, one of the things I did say. Because by definition, good math problems are messy. For this project, our writing team created problems that don’t have one right answer. For instance, one problem asks students to generate a formula to predict which players should be on the All‑NBA 1st Team. Should they use points and rebounds as part of their formula? If so, how much weight should they give to each? And should there be a deduction for the number of turnovers a player has? All of that is up to the student, and it’s certainly possible that more than one formula would give reasonable results. (If you don’t believe me, do a search for NBA Efficiency, TENDEX, Thibodeau, VORP, or New SPM to get a sense of some formulas currently used by professional statisticians.)
A microsite with a four sample problems is available at www.discoveryeducation.com/nbamath. To see all 16 problems and to experience the NBA Math Tool, you’ll need to login to Math Techbook; sign up for a 60‑day trial at www.discoveryeducation.com/mathtechbook.
I’m ecstatic about the problems that our writing team — which includes folks who love both math and basketball, like Brenan Bardige, Ellen Clay, Chris Shore, Shauna Hedgepeth, Katie Rhee, Jen Silverman, and Jason Slowbe — has created. One of the simpler problems they’ve written, meant for middle school students, is to determine which player should take a technical free throw. It’s not a hard problem, but students get to choose which team(s) to examine and how to use free-throw data to make their choice. With the NBA Math Tool that we’ve created, which includes FTM and FTA but not FT%, one possible formula is =ROUND(100*FTM/FTA,1), which will display the free-throw percentage to one decimal place of accuracy — though there are certainly less sophisticated formulas that will get the job done, too, and students could bypass formulas entirely by using equivalent fractions.
But a different article said that the “questions may look something like” this:
Andrew Wiggins is making 49.1% of his two-point shots and 52.3% of his threes. Which shot is he more likely to make?
Actually, we would never ask a question like this in Math Techbook, either as part of this NBA project or otherwise. By the time students start working with percentages in middle school (6.RPA.3.C), we expect that they already understand how to compare decimals (4.NF.C.7). Though basic exercises are included in the service, most problems — and especially those based on NBA data — exist at a greater depth of knowledge.
But what we might do is ask students to use proportions to make a prediction.
As you know, basketball announcers and sportswriters make predictions all the time. They talk about players being “on pace” to score some number of points or to grab a certain number of rebounds. In fact, the Washington Post recently prophesied that Steph Curry will hold the NBA’s all-time three-point record before the next presidential election.
During the first part of the event at Johnson Middle School, the students set out to make a prediction:
How many assists will John Wall finish the season with?
John Wall recently set the Wizards franchise record for assists, so the context was timely.
To solve this problem, students explored the NBA Math Tool, which now resides inside Math Techbook. This tool allows students to analyze both NBA and WNBA stats. Students considered data for the Washington Wizards:
Row 6 shows that John Wall had 98 assists through 11 games. Good information, to be sure, but it led to more questions from students than answers:
- Some players on the Wizards have played 13 games. How many games have the Wizards played so far this year?
- How many games will John Wall play this year?
- How many games are in an NBA season?
Looking at team data in the NBA Math Tool, students learned that the Wizards have played 13 games so far this year. And one student knew that every NBA team plays 82 games in a season. Good info… but now what?
One approach is to set up a proportion with the equation
which yields the number of games (g) that we can expect John to play this year (69), and then the equation
can be used to find the number of assists (a) that we can expect John to record (602).
But the eighty students in the gym were sixth- and seventh-graders, and they weren’t ready for algebraic equations. Instead, they attacked the problem by noting that Wall had 98 assists through the first 13 games, so they estimated:
- He should have about 200 assists through 25 games.
- He’d have about 400 assists through 50 games.
- He’d have about 600 assists through 75 games.
- That’s 7 games shy of a full 82‑game season, and Wall should have about 50 assists in 7 games.
- So, we can expect him to finish the season with about 650 assists.
My role at the event was to lead students through the solution as a group-problem solving activity; and then, to work with them in the media center on the free-throw problem described above. It was an incredible day! I got to co-teach with Ivory Latta, point guard for the Washington Mystics:
I got to meet some incredible people, including current players, former players, and NBA executives:
But most importantly, I was finally able to let the world know about this amazing project, which my team has been working on for a year.
The NBA slogan is, This is why we play. But today I say, this is why we work: to develop rich curriculum resources that are fun, relevant, and powerful in teaching kids math.
It’s well known that 5 out of 4 people have trouble with fractions, but even the mathematically advanced may have a little trouble with this puzzle. Your challenge is simple…
Find the sum of all items in the following table.
A hint is below the table, and the answers are below that. Good luck!
Hmm… it seems that you scrolled down here a little too quickly for the hint. Try harder. To put some distance between you and the hint, here are some fraction jokes:
How is sex like a fraction?
It’s improper for the larger one to be on top.
Which king invented fractions?
Henry the Eighth.
There’s a fine line between the numerator and denominator.
(And it’s called a vinculum.)
Okay, you’ve waited long enough. Here’s your hint. The items in the table are a fird (fish + bird), wooden forts, bottles of whiskey, the Sith lord Darth Maul, wraiths, and tents. (By the way, thanks to www.HikingArtist.com for the cool drawing of the fird!) Hope that helps.
To put some space between the hint and the answer, here are some more fraction jokes:
A student once told me, “To prove to you that I understand equivalent fractions, I only did three-sevenths of my homework.”
I was scared half to death… twice.
What is one-fifth of a foot?
Okay, you’ve waited (and endured) enough. Without further adieu, the answer is 2.5. The images in the table are:
- one fird
- two forts
- four fifths
- one Sith
- four wraiths
- two tents
the sum of which is
Thanks for stopping by. Have a great day!
It’s hard to believe that Who Wants to Be a Millionaire has been on the air since 1999, isn’t it? Even harder to believe is the number of math questions that have been missed by contestants.
In this post, I’m going to share five questions that have appeared on WWTBAM, followed by a brief discussion. If you’d like to solve them before reading the discussion, or if you want to share the quiz with friends or students, you can download it:
Three of the five questions were answered incorrectly by contestants. In one case, the contestant polled the audience and received some bad advice. If I hadn’t put this collection together, I’m not sure I would’ve been able to identify which ones were answered correctly. So maybe that’s a bonus question for you: Which two questions were answered correctly?
I’m unquestionably biased, but I always feel like the math questions on WWTBAM are easier than questions from other disciplines. Then again, maybe a history major would think that questions about Eleanor of Aquitaine are trivial. But take these non-math questions:
- In the children’s book series, where is Paddington Bear originally from? (Wait… he’s not from England?)
- What letter must appear at the beginning of the registration number of all non-military aircraft in the U.S.? (Like most things, it’s obvious — once you know the answer.)
- For ordering his favorite beverages on demand, LBJ had four buttons installed in the Oval Office labeled “coffee,” “tea,” “Coke,” and what? (Hint: the drink wasn’t available when he was Vice President.)
My conjecture is that non-math questions generally have an answer that you either know or don’t know, but math questions can be solved if given enough time to apply some logic and computation.
Perhaps you’ll disagree after attempting these questions.
1. What is the minimum number of six-packs one would need to buy in order to put “99 bottles of beer on the wall”?
2. Which of these square numbers also happens to be the sum of two smaller square numbers?
3. If a euro is worth $1.50, five euros is worth what?
- Thirty quarters
- Fifty dimes
- Seventy nickels
- Ninety pennies
4. How much daylight is there on a day when the sunrise is at 7:14 a.m. and the sunset is at 5:11 p.m.?
- 9 hours, 3 minutes
- 8 hours, 37 minutes
- 9 hours, 57 minutes
- 8 hours, 7 minutes
5. In the year she turned 114, the world’s oldest living person, Misao Okawa of Japan, accomplished the rare feat of having lived for how long?
- 50,000 days
- 10,000 weeks
- 2,000 months
- 1 million hours
Discussion and Answers
1. Okay, really? Since 16 × 6 = 96, one would need 17 six-packs, B.
2. This is the one for which the contestant asked the audience. That was a bad move… 50% of the audience chose A, but only 30% chose the correct answer. Since 25 = 9 + 16, and both 9 and 16 are square numbers (9 = 32, 16 = 42), the correct answer is B.
3. It’s pretty easy to calculate $1.50 × 5 = $7.50. The hard part is figuring out which coin combination is also equal to $7.50. Okay, it’s not that hard… but it took Patricia Heaton a lifeline and more than 4 minutes.
4. My question is whether daylight is officially defined as the time from sunrise to sunset. Apparently, it is. That makes this one rather easy. From 7 a.m. to 5 p.m. is 10 hours, and since 11 and 14 only differ by 3 minutes, we need a time that is 3 minutes less than 10 hours: C, final answer.
5. Without a doubt, this is the hardest of the five questions. Contestants aren’t allowed to use calculators, so they need to rely on mental math. Estimates will do wonders in this case.
- Days: 114 years × 365.25 days/year ≈ 100 × 400 = 40,000 days
- Weeks: 114 years × 52.18 weeks/year ≈ 120 × 50 = 6,000 weeks
- Months: 114 years × 12 months/year < 120 × 12 = 1,440 months
- Hours: 114 years × 365.25 days × 24 hr/day ≈ 40,000 × 25 = 1,000,000 hours
Only the last result is close enough to be reasonable, so the answer must be D.
What’s amusing is that the contestant got the correct answer, but for the wrong reasons. For instance, he estimated the number of weeks to be 50,000, not 5,000. He then used that result to say, “It can’t be 50,000 days, because it’s about 50,000 weeks.” That’s using a false premise to arrive at a correct conclusion. On the other hand, I wonder how well I’d be able to calculate in front of a national audience with $25,000 on the line. Regardless of how he got there, he correctly chose D, to which host Terry Crews said, “You took your time on this. You worked it through. It’s what we all need to do in life sometimes. And that’s how you win the game!”
Should I ever become a question writer for Millionaire, I’d submit the following:
Which of the following are incorrect answers to this question?
- B, C, D
- A, B, C
- A, C, D
- A, B, D
When I was young, we spent a lot of time on highways, driving to and from our summer cottage. I’d see a Pennsylvania license plate like the one below, which at the time had five digits and one letter. Most people, I suspect, would be unimpressed. But not me. I’d say to my parents, “How cool is that license plate? If p = 26 and the cracked bell were an equal sign, it would be 23 × 26 = 598.”
My mom would respond with, “If you say so,” or a shrug. She had failed algebra in high school and would regularly and disgustedly declare, “How the hell can x = 6, when x is a letter and 6 is a number?”
My father — who dropped out of school to join the Navy at age 15 and had never taken an algebra course — would simply grunt.
Neither of them saw the beauty in numbers. I, on the other hand, couldn’t not see it. I wasn’t mad about this. I was just sad that they couldn’t share my joy.
On my commute this morning, I saw a truck with the number 12448 on the tailgate. I mentally added two symbols and formed the equation 12 × 4 = 48.
When my boss told me that he was retiring on January 4, I remarked, “What a great choice! The numbers 1, 4, and 16 are all square numbers, and 1, 4, 16 forms a geometric sequence.”
The truth is, it’s not really possible for me to look at a number — whether it’s a license plate, calendar, billboard, identification card, lottery number, bar code, serial number, road sign, odometer, checking account, confirmation number, credit card, phone number, phone bill, receipt total, frequent flyer number, VIN, TIN, PIN, ISBN, or any of a million other numbers — and not try to figure out some way to give it meaning beyond just its digits.
I’m not the only one with this affliction. All mathy folks have mathopia — a visual disorder that causes us to see the all things through a mathematical lens.
G. H. Hardy had mathopia. He looked for a special omen in 1729, the number of the taxicab he took to visit his sick friend Srinivasa Ramanujan. Upon arriving, he mentioned that he hoped it wasn’t a bad omen to have taken a cab with such a dull number. Ramanujan had mathopia, too. He replied that 1729 was actually “an interesting number; it is the smallest number expressible as the sum of two cubes in two different ways.”
Jason Padgett, whose latent mathematical powers suddenly appeared after he sustained a brain injury, has mathopia. He explained how he sees the world:
I watch the cream stirred into the brew. The perfect spiral is an important shape to me. It’s a fractal. Suddenly, it’s not just my morning cup of joe; it’s geometry speaking to me.
This is the way that math people work. We see numbers and patterns everywhere, sometimes even when they’re not really there. Or, maybe, when they’re not meant to be there. And while I am not trying to imply that I’m anything close to Hardy or Ramanujan or Padgett, I do think that they and I shared one characteristic — the burden, and the blessing, of seeing the world through math-colored glasses.
World Sight Day, celebrated on the second Thursday of October each year — in other words, today — seems like a good day to bring awareness of mathopia to the masses. It doesn’t hurt that today is 10/13/16, a date forming an arithmetic sequence, in which all three numbers are Belgian-1 numbers. (See, I can’t turn it off.)
Do you have mathopia? What do you see when you encounter a number?
No, the title of this post does not refer to the beer. Though it may be the most interesting blog post in the world.
It refers to the date, 10/10, which — at least this year — is the second day of National Metric Week. It would also be written in Roman numerals as X/X, hence the title of this post.
For today, I have not one, not two, but three puzzles for you. I’m providing them to you well in advance of October 10, though, in case you’re one of those clever types who wants to use these puzzles on the actual date… this will give you time to plan.
The first is a garden-variety math problem based on the date (including the year).
Today is 10/10/16. What is the area of a triangle whose three sides measure 10 cm, 10 cm, and 16 cm?
Hint: A triangle appearing in an analogous problem exactly four years ago would have had the same area.
The next two puzzles may be a little more fun for the less mathy among us — though I’m not sure that any such people read this blog.
Create a list of words, the first with 2 letters, the second with 3 letters, and so on, continuing as long as you can, where each word ends with the letter X. Scoring is triangular: Add the number of letters in all the words that you create until your first omission. For instance, if you got words with 2, 3, 4, 5, and 8 letters, then your score would be 2 + 3 + 4 + 5 = 14; you wouldn’t get credit for the 8-letter word since you hadn’t found any 6- or 7-letter words.
2 letters: _________________________
3 letters: _________________________
4 letters: _________________________
5 letters: _________________________
6 letters: _________________________
7 letters: _________________________
8 letters: _________________________
9 letters: _________________________
10 letters: _________________________
11 letters: _________________________
12 letters: _________________________
13 letters: _________________________
14 letters: _________________________
Note: There are answer blanks above for words up to 14 letters, because — you guessed it — the longest English word that ends with an X contains 14 letters.
The third and final puzzle is a variation on the second.
How many words can you think of that contain the letter X twice? (Zoiks!) Scoring: Ten points for the first one, and a bazillion points for each one thereafter — this is hard! Good luck!
If you’re in desperate need of help, you can access my list of words for both puzzles — of which I’m fairly proud, since my list of words that end in X include math words for 2 through 10 letters — or do a search at www.morewords.com.
Eight days a week…
Yes, I know that this series is called A Week of KenKen, and I’m fully aware that there are only seven days in a week. But if the Beatles can love you for an extra day, then I can certainly write an extra post about KenKen. In case you’ve missed the fun we’ve had previously…
- Day 1: Introduction
- Day 2: The KENtathlon
- Day 3: KenKen Times
- Day 4: My KenKen Puzzles
- Day 5: Harold Reiter’s Puzzles
- Day 6: KenKen Glossary
- Day 7: KenKen Puzzle for 2016
Tetsuya Miyamoto created KenKen in 2004. Twelve years later, millions of KenKen puzzles are solved every day by people all over the world.
His original intent was not to create a global math sensation. Instead, he wanted to help his students improve their calculation skills, logical thinking, and persistence. Who knew that he would accomplish both?
KenKen puzzles are perfect for the classroom because they provide the same level of practice and repetition — sometimes affectionately known as drill-and-kill — as a worksheet full of problems, yet providing a significantly higher level of engagement.
Most students would have no more interest in answering the following questions than they would in removing their toenails with a pair of pliers:
Use only the numbers 1-5, in how many ways can you…
- write 300 as a product of 4 factors?
- write 40 as a product of 4 factors?
- write 13 as a sum of 5 numbers?
- write 12 as a sum of 4 numbers?
- write 5 as a quotient of 2 numbers?
Yet wouldn’t students be willing to at least try this 5 × 5 KenKen puzzle? The cognitive demand is the same, but as any marketing guru or parent trying to get their kids to eat vegetables will tell you, it’s all about the presentation.
Because of the puzzle’s appeal and impact for students, the KenKen in the Classroom program was created. Every Friday, teachers who’ve signed up will receive free puzzles, which can be printed for distribution to students.
KenKen puzzles deal with a lot of mathematics beyond the four binary operations, including factors, parity, symmetry, modular arithmetic, congruence, isomorphism, algebraic thinking, and problem solving. Harold Reiter, John Thornton, and I wrote about these topics and how to use KenKen in a secondary classroom in the article Using KenKen to Build Reasoning Skills.
Even better than solving KenKen puzzles, though, is having students design their own. And to that very point… the 5 × 5 puzzle that appears above was created by my son Alex when he was 6 years old.
I hope you’ve enjoyed reading this week of KenKen posts as much as I’ve enjoyed writing them. I’d love to hear your opinion of the series. Definitely check out the other items in this series with the links at the top of this post, and share your thoughts on all of them in the comments.
Good day, and welcome to Day 7 of the A Week of KenKen series. If you’ve stumbled onto this page randomly, you should definitely check out some of the fun we’ve had previously…
- Day 1: Introduction
- Day 2: The KENtathlon
- Day 3: KenKen Times
- Day 4: My KenKen Puzzles
- Day 5: Harold Reiter’s Puzzles
- Day 6: KenKen Glossary
Well, here it is, September 25, a mere 269 days into the year, and I am only now presenting you with a KenKen puzzle based on the year. I suppose I can take some solace in the fact that I’m presenting the following 2016 KenKen puzzle before the year is over. Little victories.
You might note that the puzzle has the following attributes:
- The numbers 2, 0, 1, and 6 are featured prominently as individual cells.
- Every target number uses (some combination of) the digits 2, 0, 1, and 6.
- There are two large cages — one in the shape of a 1, the other in the shape of a 6 — each with a product of 2016.
This puzzle follows the standard rules of KenKen, but there is one major exception: although it is an 8 × 8 puzzle, instead of using the digits 1‑8, it uses 0‑7.
If you’re stuck, just fill in the squares randomly. There are only 108,776,032,459,082,956,800 different Latin squares of size 8 × 8, and your chances of guessing correctly are even better since 4 cells are already filled in.
As it typical of all puzzles presented on this blog, I am not posting the solution. At least, not yet. Maybe someday. If you beg. Or send me money. But not today.