It’s About Time
I know, I know. It’s been a really long time since my last post. Nearly six months ago — February 25, to be exact.
I’ve got a good excuse, though. I took a new job, and I moved across the country. (More about that later.)
For now, I’m going to ease back into this with a simple post full of jokes. And I know what you’re thinking: “It’s about time!” So in honor of you, here are a collection of math jokes about time.
Did you hear about the hungry clock?
It went back four seconds.I lost my job at the calendar factory. My boss was mad that I took a few days off!
Mondays are an horrendous way to spend 1/7 of your life.
Traditional calendars are for the week-minded.
Did you hear about the two thieves who broke into a house and stole a calendar?
They each got 6 months.A broken clock is still correct twice every day.
The problem with calendars? In one year, out the other.
What’s the difference between a mathematician and a calendar?
The calendar has dates.The scientist dropped a watch into a beaker. She was hoping for a timely solution.
What did the hour hand say to the minute hand?
“Don’t listen to that other guy. He’s got second-hand information.”A calendar doesn’t feel well and visits the doctor. The doctor tells him, “I’ve got some bad news for you. You’ve got 12 months.”
My calendar was printed upside down. It was an interesting turn of events.
Did you hear about the calendar who owed money to a mobster?
His days are numbered!What type of candy never arrives on time?
Choco-late.When I was young, we were so poor that I had to use old calendar pages to wipe after defecating.
The worst days are behind me.What is a calendar’s favorite fruit?
Dates.How many months have 28 days?
All of them.How many seconds are in a year?
12: January 2, February 2, March 2, …
Okay, for reals regarding that last one. In a 365-day, non-leap year, there are 31,536,000 seconds. That’s kind of a fun number, because its prime factorization is…
…and the only digits in the prime factorization are the four single-digit primes. Cool stuff.
When Math Falls into the Wrong Hands
My brother-in-law recently forwarded an email that contained a lot of images plucked from various degenerate corners of the internet, and he suggested that this one could go into my next book:
I suppose it’s funny enough, and I guess it’s technically a math joke, but there’s a problem.
It doesn’t work.
I know, I know. Most people just read the joke, get the humor that the note’s author has used some odious expression to represent the PIN code, and go on about their day. Plus, I’ve heard that less than 1% of the world’s population has taken calculus, so there aren’t too many people who could actually check the math. Not to mention, how many of them would care enough to do so?
Uh… I can think of at least one person who cares enough.
While it’s certainly egotistical to think that I’m the only one in the intersection, it’s likely offensive to include anyone in the intersection who really wouldn’t want to be. So apologies to Matt Parker, Des McHale, Colin Adams, Ed Burger, or any of the other funny math folks who think they should have been included.
Anyway, where was I? Oh, right. Bad math.
The definite integral in the joke sent by my brother-in-law doesn’t yield a four-digit positive integer.
In fact, it yields a very irrational number with a lot of digits:
-2.58208625277854512796640677001459519299166472798789689499…
So unless the PIN code for that bank card has an infinity of digits, well, this is going to be problematic.
I propose, instead, that the joke be rewritten to use the following:
Would it be less funny? Probably. But at least it’d be accurate.
Not to mention, it would be a significantly more fair to Darling. Honestly, no one should ever have to do integration by substitution.
The Other Golden Ratio
You’re a math geek. I know that to be true, because you’re reading this blog. And I also know that when you hear golden ratio, you think of this:
Or this:
But there’s another, different — and, dare I say, better? — golden ratio that may be even more important to learn. Especially if you’re one of the mathy folks to whom this adage applies:
Wherever you find four mathematicians, you’ll likely find a fifth.
As Homer Simpson says, “It’s funny ’cause it’s true.”
Like many classroom teachers, I’m often ready for a cocktail on Friday afternoons. And like those teachers, I don’t want to spend a lot of time thinking about it; I don’t want to rummage for a recipe; I just want to relax and have a drink. But, I also don’t want to have the same cocktail every Friday; variety is the spice of life.
So, what’s a boy to do?
Simple. Follow the advice from the folks in the Food Hacks division at Wonder How To, who claim that the following ratio will yield a delicious cocktail every time:
2 : 1 : 1 :: alcohol : sour : sweet
Right now, you’re probably scratching your head and thinking, “Can it really be that simple?”
I’m here to tell you, friends — it is.
For instance, 2 parts tequila, 1 part lime juice, and 1 part triple sec? That’ll get you a tasty margarita.
And 2 parts bourbon, 1 part lemon juice, 1 part simple syrup? None other than a classic whiskey sour.
If you combine 2 parts tequila, 1 part rhubarb liqueur, and 1 part malic acid — it’s what gives green apples their tartness — then you’d be getting close to a drink they call The Scarlet Lantern at Bar Congress in Austin, TX. (They also mix in black cardamom-strawberry shrub. Keep Austin weird, eh?)
Now that you know about the other golden ratio, here’s what you need to do: Organize your liquor cabinet into two parts, hard alcohol and sweet mixers. Then, make sure you keep a couple of sour mixers in your fridge. When you get home on Friday, just grab a bottle from each side of the liquor cabinet and one more from the fridge, pour, and — voila — instant happiness.
Wanna get a little crazy? Find your favorite cube-shaped random number generator, give it three rolls, then choose the appropriate item from each column in the table below.
Die Roll |
Alcohol |
Sour |
Sweet |
1 |
Tequila |
Lemon Juice |
Simple Syrup |
2 |
Vodka |
Lime Juice |
Triple Sec |
3 |
Rum |
Grapefruit Juice |
Cointreau |
4 |
Rye Whiskey |
Strawberry Shrub |
Gran Marnier |
5 |
Bourbon |
Bloody Mary Mix |
Honey-Ginger Syrup |
6 |
Mezcal |
Dry Cider |
Grenadine |
Personally, I’m hoping for 6-3-5, which is kind of like a Mezcal Paloma, sort of like a Honey and Smoke, but not really similar enough to be either one. So, I guess I get to name it. And given what I’ve heard about the fat-burning properties of honey and grapefruit juice, I’m going to call this newly minted beverage the Weight Watcher.
Now we just need to come up with names for the other 215 combinations. I’ll get started on that right away… soon as I finish this drink.
Two the Hard Way (and an Easy Way)
During our trip to Arizona for winter break, two problems surfaced organically while we were on holiday. (Sorry. Two math problems. There were lots of non-math problems, too, but we don’t have time for all that.)
On the plane, the interactive in-flight map showed the outside temp, toggling between Fahrenheit and Celsius. That led to the following MJ4MF original problem, which I thought — and still think — is pretty good:
The conversion between Fahrenheit and Celsius temperatures follows the rule F = 9/5 C + 32. Sometimes, the temperature is positive for both Celsius and Fahrenheit; sometimes, the temperature is negative for both Celsius and Fahrenheit; and other times, Fahrenheit is positive while Celsius is negative. What is the least possible product of the Fahrenheit temperature and its corresponding Celsius temperature?
You can pause here if you’d like to solve this before I present a spoiler.
Before I reveal two different solutions, allow me to digress. It could be that the statement about the temps sometimes being positive and sometimes being negative is denying a teachable moment. The graph below shows the linear relationship between the two temperature scales. Perhaps a good classroom question is:
When will the product CF be positive and when will it be negative?
Or maybe a better question is:
When are C and F both positive, when are they both negative, and when do they have different signs?
So, to the problem that I posed. As I thought about it on the plane, I concluded that if F = 1.8C + 32, then the product CF = 1.8C^{2} + 32C. I then used calculus, found the derivative (CF)’ = 3.6C + 32, set that equal to 0, and concluded that C = ‑8.89, approximately. The corresponding Fahrenheit temperature is F = 16, so the minimum product is roughly ‑142.22.
Using calculus was like rolling a pie crust with a steamroller, though. I could have just as easily graphed the parabola and noted its vertex:
If I had used the other form of the rule, namely C = 5/9 (F ‑ 32), things might have been a little easier. Maybe. In that case, setting the derivative equal to 0 yields F = 16, which is arguably a nicer number. But then you still have to find the corresponding Celsius temperature, which is C = ‑8.89, and the product is still roughly ‑142.22. So, not much easier, if at all, and again graphing the parabola and noting its vertex would have done the trick:
The only real benefit to using this alternate version of the rule is that it provides a reasonable check. Since both methods — and both graphs — yield an answer of ‑142.22, we can feel confident in the result.
But there’s an easier way to solve this one.
Thinking this was a good problem — and because I like when my sons make me feel stupid — I gave it to Eli and Alex. Within seconds, Eli said, “Well, F is positive and C is negative between 0°F and 32°F, so the minimum will occur halfway between them at F = 16. That means C = ‑80/9, so it’s whatever ‑1280/9 reduces to.” (Turns out, -1280/9 = ‑142 2/9 ≈ ‑142.22.)
Eli hasn’t taken calculus, so he doesn’t know — or, at least, he hasn’t learned — that the minimum product should occur halfway between the x– and y‑intercepts of the linear graph. Yet, he had an intuitive insight that just happens to be true. As a result, what took me about five minutes of deriving and manipulating took him about five seconds.
The second problem arose at the grocery store. Among our purchases was a box of sugar cubes, which contained, surprisingly, 126 cubes.This number is surprising in the sense that it’s not a number you’ll see very often, except for an occasional appearance in the ninth row of Pascal’s Triangle, or maybe if you’re a chemist searching for stable atoms.
A question that could have arisen from this situation involves surface area and volume:
A rectangular prism with integer dimensions has a volume of 126 cubic units. What is the least possible surface area?
That’s not the question that was shared with Alex and Eli, though. (The answer, if you care, is 162 square units, which results from a 3 × 6 × 7 arrangement — which, in fact, is the exact arrangement of cubes in the box above. I suspect this is not a coincidence.)
The problem that I shared with my sons involved probability:
Imagine that the arrangement of cubes is removed from the box intact, and all six faces of the prism are painted red. Then one of the sugar cubes is selected at random and rolled. What is the probability that the top face of the rolled cube will be red?
The boys made an organized list, as follows:
Painted Faces | Number of Cubes |
3 | 8 |
2 | 40 |
1 | 58 |
0 | 20 |
Further, the boys reasoned:
- P(cube with 3 red faces, red face lands on top) = 8/126 x 1/2 = 8/252
- P(cube with 2 red faces, red face lands on top) = 40/126 x 1/3 = 40/378
- P(cube with 1 red face, red face lands on top) = 58/126 x 1/6 = 58/756
Therefore,
- (P of getting a red face) = 8/252 + 40/378 + 58/576 = (24 + 80 + 58) / 576 = 162/576 = 9/42
Wow! That seems like a lot of work to get to the answer. Surely there’s an easier way, right?
Indeed, there is.
Notice that the penultimate step yielded the fraction 162/576. The numerator, 162, may look familiar. It’s the answer to the question that wasn’t asked above, the one about the least possible surface area of the prism. That’s no coincidence. In total, there will be 162 faces painted red. And there are 6 × 126 = 576 total faces on all of the sugar cubes (that is, six faces on each cube). This again suggests that the probability of rolling a red face is 162/576.
Did you happen to notice that the volume and surface area use the same digits in a different order? Cool.
So there you have it, two problems, each with two solutions, one easy and one hard. Or as mathematicians might say, one elegant and one common.
It’s typical for problems, especially problems worth solving, to have more than one solution strategy. What’s the trick to finding the elegant solution? Sadly, no such trick exists. Becoming a better problem solver is just like everything else in life; your skills improve with practice and experience. It’s akin to Peter Sagal’s advice in The Incomplete Book of Running, where he says, “You want to be a writer? […] Just sit down and write. The more you write, the better a writer you will become. You want to be a runner? Run when you can and where you can. Increase your mileage gradually, and your body will respond and you’ll find yourself running farther and faster than you ever thought possible.” You want to be a problem solver? Then spend your time solving problems. That’s the only way to increase the likelihood that you’ll occasionally stumble on an easy, elegant solution.
And every once in a while, you may even solve a problem faster than your kids.
Silent Letter Night
Several weeks ago, Will Shortz presented an NPR Sunday Puzzle in which he stated a word and a letter, and the resulting collection would be rearranged to form a new word in which the added letter is silent. For instance, if Will gave RODS + W, the correct answer would be SWORD, in which the W is silent. (Note that the collection of letters is also an anagram of WORDS, but the W isn’t silent.)
At a time of year known for silent nights, it seems like a puzzle involving silent letters is completely appropriate. I’ve borrowed Shortz’s idea and extended it a bit; some of the clues in the list below have more than one silent letter added. Many items in the list are related to today’s holiday; and, because this is a math blog, the others are related to mathematics. In full disclosure, two of the answers are proper nouns.
Enjoy, and happy holidays!
- TO + W =
- TON + K =
- TOGS + H =
- GEE + D =
- TIN + G + H =
- SIN + G =
- SIN + E =
- CORD + H =
- HOLE + W =
- HEART + W =
- TINNY + E =
- COINS + E =
- NOELS + M =
- PILES + E + L =
- FRAME + T =
- REDACT + E + S + S =
- RACISMS + H + T =
12 Math Games, Puzzles, and Problems for Your Holiday Car Trip
It doesn’t matter if you’re one of the 102.1 million people traveling by car, one of the 6.7 million people traveling by plane, or one of the 3.7 million people traveling by train, bus, or cruise ship this holiday season — the following collection of games, puzzles, and problems will help to pass the time, and you’ll be there long before anyone asks, “Are we there yet?”
Games
1. Street Sign Bingo
Use the numbers on street signs to create expressions with specific values. For instance, let’s say you see the following sign:
The two numbers on the sign are 12 and 3, from which you could make the following:
12 + 3 = 15
12 – 3 = 9
12 × 3 = 36
12 ÷ 3 = 4
Of course, you could just use the numbers directly as 3 or 12. Or if your passengers know some advanced math, they could use square roots, exponents, and more to create other values.
Can you split the digits within a number and use them separately? Can you concatenate two single-digit numbers to make a double-digit number? That’s for you and your traveling companions to decide.
To play as a competitive game, have each person in the car try to get every value from 1‑20. It’s easiest to keep track if you require that players go in order. To make it a cooperative game, have everyone in the car work collectively to make every value from 1 to 100.
As an alternative, you could use the numbers on license plates instead.
2. Bizz Buzz Bang
Yes, I’ve played this as a drinking game. No, I don’t condone drinking while driving. No, I don’t condone under-age drinking, either. Yes, I condone playing this game with minors while driving. (See what I did there?)
The idea is simple. You pick two single-digit numbers, A and B. Then you and your friends start counting, one number per person. But each time someone gets to a number that contains the digit A or is a multiple of A, she says, “Bizz!” Every time someone gets to a number that contains the digit B or is a multiple of B, she says, “Buzz!” And every time someone gets to a number that meets both criteria, he says, “Bang!”
For example, let’s say A = 2 and B = 3. Then the counting would go like this:
one, bizz, buzz, bizz, five, bang, seven, bizz, buzz, bizz, eleven, bang, buzz, …
When someone makes a mistake, that round ends. Start again, and see if you can beat your record.
You can use whatever numbers you like, or modify the rules in other ways. For instance, what if bizz is for prime numbers and buzz is for numbers of the form 4n + 3? Could be fun!
3. Dollar Nim (or any other variation)
Nim is a math strategy game in which players take turns removing coins from a pile. Different versions of the game are created by adjusting the number of coins that can be removed on each turn, the number of coins in the pile originally, how many piles there are, and whether you win (normal) or lose (misère) by taking the last coin.
A good aspect of Nim is that you don’t actually need coins. To play in the car, players just need to keep track of a running total in their heads.
A version of Nim dubbed 21 Flags was played on Survivor Thailand several years ago. There were 21 flags, and each team could remove 1, 2, or 3 flags on each turn. The team to remove the last flag won. This is a good first version to play in the car, especially for young kids who have never played before. Then mix it up by changing the initial amount and the number that can be removed on each turn.
Our family’s favorite version, Dollar Nim, is played by starting with $1.00 and removing the value of a common coin (quarter, dime, nickel, penny) on each turn. (While discussing this post with my sons, they informed me that they much prefer Euro Nim, which begins with 1€, but then the coin values to be removed are 50c, 20c, 10c, 5c, 2c, and 1c. It’s essentially the same game, but they like the different coin amounts.)
Three-Pile Nim consists of not just one but three piles with 3, 5, and 7 coins, respectively. On each turn, a player must remove at least one coin, and may remove any number of coins, but all removed coins must be from the same pile.
Finally, Doubling Nim is played as the name implies. On the first turn, a player may remove any number of coins but not the entire pile. On every turn thereafter, a player may remove any number of coins up to double the number taken on the previous turn. For instance, if your opponent removes 7 coins, then you can remove up to 14 coins.
For every version of Nim, there is an optimal strategy. We’ve wasted hours on car trips discussing the strategy for just one variation. Discussing the strategies for all the versions above could occupy the entire drive from Paducah to Flint.
4. Guess My Number
One person picks a number, others ask questions to try to guess the number.
The simplest version is using “greater than” and “less than” questions. Is it greater than 50? Is it less than 175? And so forth. Using this method, the guessers can reduce the number of possibilities by half with each question, so at most, it should take no more than n guesses if 2^{n} > m, where m is the maximum possible number that the picker may choose. For instance, if the picker is required to choose a number less than 100, then n = 7, because 2^{7} = 128 > 100. Truthfully, this version of the game gets boring quickly, but it’s worth playing once or twice, especially if there’s one picker and multiple guessers. And a conversation about the maximum number of guesses need can be a fun, mathy way to spend 15 minutes of your trip.
A more advanced version excludes “greater than” and “less than” questions. Instead, the guessers can ask other mathematical questions like, “Is it a prime number?” or “Do the digits of the number differ by 4?” Those questions imply, of course, that the answer must be yes or no, and that’s typical for these types of guessing games. If you remove that restriction, though, then guessers could ask questions that reveal a little more information, like, “What is the difference between the digits?” or “What is the remainder when the number is divided by 6?” With this variant, it’s often possible to identify the number with two strategic questions.
Puzzles
Here are three puzzles that can lead to hours of conversation — and frustration! — on a car trip. Before you offer one to your crew, though, put forth the disclaimer that anyone who’s heard the puzzle before must remain mum. No reason they should spoil the fun for the rest of you. (The puzzles are presented here without solution, because you’ll know when you get the right answer. You can find the answer to any of them online with a quick search… but don’t do that. You’ll feel much better if you solve it yourself.)
5. Dangerous Crossing
Four people come to a river in the middle of the night. There’s a narrow bridge, but it’s old and rickety and can only hold two people at a time. They have just one flashlight and, because it’s night and the bridge is in disrepair, the flashlight must be used when crossing the bridge. Aakash can cross the bridge in 1 minute, Britney in 2 minutes, Cedric in 5 minutes, and Deng in 8 minutes. When two people cross the bridge together, they must travel at the slower person’s pace. And they need to hurry, because zombies are approaching. (Oh, sorry, had I failed to mention the zombie apocalypse?) What is the least amount of time that all four people can cross the bridge?
And how can you be sure that your method is the fastest?
6. Weight of Weights
Marilyn has a simple balance scale and four small weights, each weighing a whole number of grams. With the balance scale and these weights, she is able to determine the weight of any object that weighs between 1 kg and 40 kg. How much does each of the four weights weigh?
7. Product Values
Assign each letter a value equal to its position in the alphabet, i.e., A = 1, B = 2, C = 3, …, Z = 26. Then for any common word, find its product value by multiplying the value of the letters in the word. For instance, the product value of CAT is 60, because 3 × 1 × 20 = 60.
- Find as many common English words as you can with a product value of 60.
- Find a common English word that has the same product value as your name. (A little tougher.)
- Find a common English word with a product value of 3,000,000. (Zoiks!)
Problems
8. The Three of Life
This one looks so innocent!
What’s the probability that a randomly chosen number will contain the digit 3?
But spend a little time with it. And prepare for. Mind. Blown.
9. Hip to Be (Almost) Square
No calculators for this one.
There are four positive numbers — 1, 3, 8, and x — such that the product of any two of them is one less than a square number. What is the least possible value of x?
No spreadsheets, either.
10. Hip to Be Square (Roots)
Just some good, old-fashioned algebra and logic to tackle this one.
Which of the following expressions has a greater value?
or
Surprised?
11. Coming and Going
Potentially counterintuitive.
At the holidays, Leo drove to his grandma’s, and the traffic was awful! His average speed was 42 miles per hour. After the holidays, however, he drove home along the same route, and his average speed was 56 miles per hour. What was his average speed for the entire trip?
And, no, your first guess was most likely not correct.
12. The Year in Numbers
A moldy oldie, to be sure, but this puzzle is always a crowd-pleaser for those who haven’t seen it before.
Use the digits of the new year — 2, 0, 1, and 9 — and any mathematical operations to form the integers from 1 to 100. For instance, you can form 1 as follows: 2 × 0 × 9 + 1 = 1.
For an added challenge, add the restriction that you must use the four digits in order.
Bonus: A Book
My sons get sick when they read in the car. That’s why I love the two books Without Words and More Without Words by James Tanton. Literally, there are no words! Each puzzle is presented using a few examples, and then students must follow the same rules to solve a few similar, but more challenging, puzzles.
Wherever you’re headed during the holiday break — driving to your relatives’ house, flying to Fort Lauderdale, or just relaxing at home — I hope your holidays are filled with joy, happiness, and lots of math!
Where Would We Fit?
In the classic book 101 Puzzle Problems, Nathaniel B. Bates and Sanderson M. Smith make an astounding claim:
The volume of the 1970 world population is less than the volume of the Houston Astrodome.
That is, if you packed all of the people on Earth who were alive in 1970 — all 4 billion of them — as tightly as possible, they would’ve actually fit in the stadium, and they could’ve seen Evel Knievel jump 13 cars or Billy Jean King defeat Bobby Riggs.
Before you read any farther, you might wish to do a Fermi estimate to verify the reasonableness of this claim.
It’s an amazing fact!
Unfortunately, it’s also wrong.
Bates and Smith argue that the average human body has a volume of approximately 2 cubic feet. Their argument? That water weighs 62.5 pounds per cubic foot, and “the human body consists mostly of water.” Fair enough, if we think that an average human weighs about 125 pounds, which they probably didn’t, even in the 1970s.) Personally, I would’ve attacked the estimate a little differently. An average human can fit in a box that is 6 feet tall with a base that measures approximately 0.5 square feet — I know; I’ve been to funerals — and such a box has a volume of 3 cubic feet. But there’s a fair amount of wasted space in that box so, yeah, 2 cubic feet seems like a reasonable estimate.
Therefore, the world population in 1970 would have had a combined volume of 8,000,000,000 cubic feet.
So far, so good.
But Bates and Smith then make their statement about the Astrodome, with no calculations or estimates about the volume of the stadium to justify the claim. Please understand, I don’t fault Bates and Smith; as I understand it, the claim about everyone fitting in the Astrodome was a ubiquitous factoid during the decade of bellbottoms and disco. Still, you’d think that two mathematicians would have checked the stat before including it in a book.
The Astrodome has an outside diameter of 710 feet and a height of 208 feet. A cylinder with those dimensions would have a volume of
V = π × 355^{2} × 208 ≈ 82,000,000 cubic feet
Although this is surely an overestimate, it’s still an order of magnitude too small to fit the world’s population in 1970. There’s not nearly enough space to store all those buggers.
Bates and Smith state that the volume of the 1970 population “takes up about one-eighteenth of a cubic mile.” And I think therein lies the error. The height of the Astrodome is just shy of one-eighteenth of a mile, and the diameter is a whole lot more than one-eighteenth of a mile, so someone (incorrectly) assumed that a container with those dimensions would easily hold 1/18 cubic mile. But that’s not right. As any geometry student will tell you, shrinking each dimension to a fraction of its original length results in a volume that is the cube of that fraction. Oops.
The diameter of the Astrodome is about 1/7 of a mile, and the height is about 1/22 of a mile, so its volume would have been less than 1/1000 of a cubic mile.
So, what container would have been big enough to hold the 1970 population?
Honestly, who cares?
Things have changed. Today, we are bumping up against 8 billion people worldwide, the Astrodome was declared “unfit for occupancy” in 2016, and instead of calculating the volume of stadia, we can simply use Google to find the volume of some really large places.
In fact, Google claims that the volume of the Astrodome is 42,000,000 cubic feet. If we can believe what we read, then the calculations above were, sadly, unnecessary.
So, you may be wondering, what places in the world could hold all of us? That is, what has a volume of 16,000,000,000 cubic feet?
The building with the greatest volume in the world is the Boeing Everett Factory. It has a volume of almost 500,000,000 cubic feet — which means you’d need 32 of them to fit today’s world population.
The manmade structure with the greatest volume is the Great Wall of China, tipping the scales at just under 20,000,000,000 cubic feet. That would be large enough, though how would you get all of those people inside? For easier packing, you could turn to nature and just dump everyone into Sydney Harbour, which has a volume of 562,000 megaliters, or about 19,500,000,000 cubic feet.