## Posts tagged ‘puzzle’

### Sidhu Moose Wala Would Love This Post

The following is a modified version of a problem that appeared in* Concept Quests*, a collection of problems originally written by Kim Markworth and soon to be offered by The Math Learning Center as a supplement to *Bridges in Mathematic Second Edition*.

**I was today years old when I realized that I’ve been alive for 47 years, 47 months, 47 weeks, and 47 days. How old will I be on my next birthday?**

Incidentally, the number 47 was also featured in this week’s challenge on the NPR Sunday Puzzle:

**Take this equation: 14 + 116 + 68 = 47. Clearly this doesn**‘**t work mathematically. But it does work in a nonmathematical way. Please explain.**

The number 47 is very interesting, especially if you’re a member of The 47 Society. But even if you aren’t, the number 47 has some noteworthy credentials:

- The number 47 is a
*safe prime*, meaning it has the form 2*q*+ 1, where*q*is also prime. - The greatest number of cubes (not necessarily the same size) that cannot be rearranged to form a larger cube is 47. That is, it’s not possible to make a cube from 47 smaller, not-necessarily-the-same-size cubes, but it is possible to make a cube from 48, 49, 50, 51, or any greater number of cubes.
- 47 = 2
^{5}+ 5^{2}– 2 × 5 - Lord Asano Takumi’s followers swore to avenge his death — a suicide he was obliged to commit for drawing his sword in the palace — and survives in a story known as the Legend of the 47 Ronin.
- Mexican revolutionary Pancho Villa was killed by 47 bullets.
- It takes 47 divisions of one cell to produce the number of cells in the human body; that is, there are approximately 2
^{47}cells in a human. - Proposition 47 of Euclid’s Elements is the Pythagorean theorem.
- The Bible credits Jesus with 47 miracles.
- The Declaration of Independence has 47 sentences.
- There are 47 strings on a concert harp.
- The tropics of Cancer and Capricorn are located 47 degrees apart.

This fascination with the number 47 seems to have started a long time ago.

In the summer of 1964, two students at Pomona College — Laurens “Laurie” Mets and Bruce Elgin — began an extracurricular experiment to determine how often the number 47 occurred in nature. While this may seem like an odd thing to do, they weren’t the only ones; a number of Pomona students were involved in a summer program sponsored by the NIH, and many of them conducted similar experiments about the occurrence of random numbers, questioning the existence of patterns in nature. However, it was Mets’s and Elgin’s search for 47 that took on a life of its own and developed a cult following.

The experiment was to determine if 47 showed up in nature more often than other numbers. As a first step, they hypothesized that 47 would appear in two percent of all California license plates, which would be higher than would occur by random chance. According to Mets, a funny thing happened: when they looked at a bunch of license plates and counted the occurrence of 47, they found their hypothesis to be true. According to Elgin, however, the license plate experiment was a failure; but, within the next week, he saw a Rolaids commercial claiming that Rolaids absorbs up to 47% of its weight in excess acid. That sparked a desire to see where else 47 might occur… and before long, people all over campus were counting everything to look for 47s.

As part of that same summer program, statistician Donald Bentley gave a talk in which he stated that any number could be shown to be equal to any other number. In his (invalid) proof, he chose to show that every number was equal to — you guessed it — 47. How did he prove it?

Consider an isosceles right triangle for which the base is 47 units, and the congruent legs are each 37 units. Now, connect the midpoints of the congruent sides to the midpoint of the base, as shown in **Figure 1**. The combined lengths of the red segments in Figure 1 is the same as the sum of the lengths of the congruent sides, 74 units. Then repeat, connecting the midpoints of these smaller triangles, as shown in **Figure 2**. Again, the combined lengths of the red segments is 74 units. Then do this again and again and again — all the way to infinity — and eventually the triangles will get so small that the red segments will appear to be a straight-line segment that overlaps the base of the triangle, as shown in **Figure 3**. The length of the red segment is 74 units, and length of the base is 47 units, so 74 = 47. By similar (incorrect) reasoning, it would be possible to show that any number is equal to 47.

Although Dr. Bentley’s proof was meant as a classroom lesson on statistical computing, leave it to the 47-hunters to accept the proof and believe that all numbers equal 47.

(If you’d like to know more about this sordid history of 47, you can read The Mystery of 47 at the Pomona College website.)

Anyway, where was I? Oh, yes… I’ve been alive for 47 years, 47 months, 47 weeks, and 47 days.

Mathematically, what I find most interesting is the general case. That is, if I’ve been alive for *n* years, *n* months, *n* weeks, and *n* days, how old will I be on my next birthday?

For the problem I presented at the top of this post, *n* = 47, and the answer is 52 years; that is, I’ll turn 52 on my next birthday. But if *n* = 48, I won’t turn 53 on my next birthday; I’ll turn 54. That’s right, it skips a year. This happens periodically, and with a little help from the Date Calculator at www.timeanddate.com, I was able to generate this beautiful — albeit incomplete — graph:

If you’re interested in seeing the data, check it out at https://www.desmos.com/calculator/u43gz6ahyi.

Something really wacky happens when n = 38. I was born on March 17, 1971, and the date 38 years, 38 months, 38 weeks, and 38 days from my date of birth was **exactly** 42 years later: March 17, 2013. Whoa.

Who knows what any of this means? Perhaps this implies that 38, not 47, is a magical number. In any case, I hope you have fun playing with this scenario and seeing what you can discover.

If that’s not your thing, though, here are some time-and-date jokes for you:

How many seconds are in a year?*Twelve. January 2nd, February 2nd, March 2nd, …*

Did you hear about the hungry clock?*It went back four seconds.*

Traditional calendars are for people who are week-minded.

I was fired from my job at the calendar factory. My boss was mad that I took a few days off.

### Year-End Puzzle

There’s a math puzzle that’s been making its rounds on the web with the following proclamation:

**82 beers minus your age plus $40 is the year you were born!**

Of course, it gets way more interesting on the web, where you’ll find images of beer:

Don’t even get me started about the grammar and punctuation. (Why is “Beers” capitalized? And “your”? Ugh.)

You’ll also find pretty colors and clip art of beer mugs:

I’m also going to ignore the nonsensical nature of misaligned units — beers, years, and dollars? WTF?

But I won’t ignore the math. When the puzzle is shared online, it’s often accompanied by expressions of amazement — “OMG! This is amazing!” — or statements about its veracity — “This *always* works!” And it’s the latter claim that’s worth investigating.

To begin, I’ll show that this equation does, in fact, hold for me. I was born in 1971, and I’m 51 years old:

82 – **51** + 40 = **71**

So far, so good.

But my friend Mike turns 51 on Friday. So he tried it:

82 – **50** + 40 = **72**

Uh-oh. He was born on December 23, 1971, not 1972. It didn’t work for Mike; in fact, it won’t work for anyone who hasn’t had their birthday this year. Luckily, it’s late December, so this only excludes about 3% of the population; but in the US alone, that’s about 9 million people.

Separately, a version of this trick was texted to my sons, who are 15‑year‑old twins. They tried it:

82 – **15** + 40 = **107**

Nope! They were born in 2007. Granted, the result of 107 means 107 years *after 1900*. But that’s inconsistent with the result being the last two digits of the year, so I’m not accepting it. In fact, this trick won’t work for anyone who wasn’t born in the 1900s. The graph of age distribution below — females on the left, males on the right — shows that approximately 90 million people were born in the year 2000 or later (age 22 or younger on the graph below). Of course, about 3% of them (3 million) were already counted above — those are the people who haven’t had birthdays this year *and* are too young for the trick to work. Which means that this trick doesn’t work for about 13 + 90 – 3 = 100 million people. And that’s just in the US.

So let’s say you know someone who *was* born in the 1900s and who *has* celebrated a birthday this year. Well, you could share this trick with them… but, you better do it soon! This trick won’t work once the calendar flips to 2023. (You could, of course, modify and start the trick with 83 beers or subtract 39, and then it’ll be fine. Though it’ll still only apply to people born in the 1900s who’ve already had their birthdays, and those people will be a little harder to find in January.)

This trick requires so many caveats, it hardly seems worth sharing, right? Well, not if you’re a math teacher and especially if you’re a secondary math teacher. Above, we dabbled in a little set theory with the principle of inclusion and exclusion while considering

(people born in the 1900s) ∪ (people who’ve had birthdays this year).

It can also lead to a good discussion of domain and range. The function *y*(*a*), where *y* is the year based on your age, *a*, can be simplified as:

*y*(*a*) = 82 – *a* + 40 = 122 – *a*

A restriction on the domain of 22 < *a* < 122 allows the trick to work just fine. And from a practical standpoint, the restriction that *a* < 122 is probably unnecessary. Lucille Randon is currently the oldest person in the world at 118 years of age, so unless you plan to share the trick with a giant tortoise or a bristlecone pine tree, you’re probably okay.

But I don’t want to leave you like that, so here’s a math trick that *will* always work.

- In the grid below, circle any number. Then, cross out the other numbers in the same row or column. (For instance, if you circled the 204 in the upper left cell, you’d then cross out 506, 402, and 305 in the top row as well as 505, 406, and 307 in the left column.)
- Circle one of the remaining nine numbers, and again cross out the other numbers in the same row or column.
- Circle one of the remaining four numbers, then cross out the other numbers in the same row or column.
- This should leave just one number. Circle it.
- At this point, one number should be circled in each row and column. If that’s not the case, start over at Step 1. If that is the case, though, good job! Find the sum of the four circled numbers.

I obviously have no idea which four numbers you selected. But I do know that the sum you obtained was **2023**. (Now would be an appropriate time for applause. But that may be a little awkward if you’re reading this at work and folks in neighboring cubicles are within earshot.)

I know, it feels like magic. But it’s not; it’s just math.

Happy holidays, and happy new year!

### The Great Puzzle Hunt

A woman after my own heart, Millie Johnson loves both puzzles and jokes. She regularly posts to Facebook, and these two gems appeared recently in her feed:

- If all whole numbers between 1 and 1000 were spelled out and arranged in alphabetical order, which number would be next-to-last?

- Arrange the digits 0‑9 into a ten-digit number such that the leftmost
*n*digits comprise a number divisible by*n*. For example, if the number is ABCDEFGHJK, the three-digit number ABC must be divisible by 3, the five-digit number ABCDE must be divisible by 5, and so on.

The answers to both can be found with an online search, and the latter can be computed in milliseconds by writing some code, but you’ll have more fun if you find the answers using that big lump of gray matter in your skull.

She also recently shared this joke, which I just adore:

Millie is also the founder and puzzle creator for the **Great Puzzle Hunt**, a free, fun, full-day, team puzzle-solving event. While folks in and around Bellingham, WA, on April 9 will be lucky enough to participate in the face-to-face event on the Western Washington University campus, the rest of the world is invited to participate virtually. There are divisions for secondary (middle and high school), open (any age), WWU students, and WWU alumni. And in case you missed the subtle mention above, I’ll say it again: registration for a team of up to six participants is FREE!

Participants will engage with four puzzles; then, the answers to those puzzles will be used to form a fifth and final meta-puzzle. How hard are the puzzles? See for yourself; the puzzle below, **My Life Is In Ruins!**, is from the 2021 Great Puzzle Hunt:

The Great Puzzle Hunt is a full-day event, so pack a lunch! But my sons and I participated last year and had a phenomenal time. By the end, we were mentally exhausted but invigorated and intellectually satisfied. If you like to solve puzzles, head on over to https://www.greatpuzzlehunt.com/register and register today!

### Math Puzzles with Letters

This week on the NPR Sunday Puzzle, host Will Shortz offered the following challenge:

Name a famous city in ten letters that contains an

s. Drop thes. Then assign the remaining nine letters their standard value in the alphabet — A = 1, B = 2, C = 3, etc. The total value of the nine letters is only 25. What city is it?

It’s not much of a spoiler to note that the average value of those nine letters must be less than three, since their sum “is only 25.” Consequently, a lot of those letters must occur at the beginning of the alphabet and — if eight of them were *a*‘s — there would be no letters later than *q* in the name of the city. But that’s as much as I’ll say; you can solve the puzzle on your own. (When you do, you can submit your answer for a chance to play next week’s on-air puzzle live with Will Shortz.)

Mathematician Harold Reiter uses a similar problem with elementary school students. Using the same idea — that each letter has a value (in cents) equal to its position in the alphabet — he asks students to find a dollar word, that is, a word whose letters have a sum of 100. As it turns out, there are many. Based on a nonexhaustive search, there are at least 3,500 dollar words, and likely a whole lot more. In a quick perusal of the list, one word jumped out: **oxygon**. Nope, that’s not a typo. It’s an archaic term meaning “a triangle with three acute angles.”

All of this talk of letters reminds me of my favorite puzzle, which I call Product Values. Using the same scheme — that is, A = 1, B = 2, C = 3, etc. — find the product value of a word by *multiplying* the values of the letters. So, for instance, *cat* has a product value of 3 × 1 × 20 = 60. How many words can you find that have a product value of 100? Based on the ENABLE word list, there are nine. (If you need some help, you can use the Product Value Calculator at www.mathjokes4mathyfolks.com.)

To end this post, a few math jokes that involve letters:

And Satan sayeth, “Let’s put the alphabet in math.” Bwa-ha-ha-ha-ha.

Romans had no trouble with algebra, because X was always equal to 10.

### Change the Vowel

The following puzzle contains a clue within each clue. The answer, of course, fits the clue, but each answer is also one of the words within the clue with its vowel sound changed. For instance, the clue “the distance from the top of your hat to the sole of your shoes” contains the word *hat*, and if you change the vowel sound from a short *a* to a long *i*, you get *height*, which fits the description.

As always, there’s a catch. Every answer to the clues below is a mathy word.

- The number of permutations of three different colored socks.

- This value is the same for 3 + 1 and 2 + 2.
- Do this with two odd numbers and you’ll get an even number.
- Three hours before noon.
- You may fail to correctly expand (
*a*+*b*)(*c*+*d*) if you don’t remember this mnemonic. - One represents this fractional portion of toes on the hoof of a deer.
- The last element of a data set arranged in descending order.
- The measure of central tendency made from the most common data points.
- The number of contestants ahead of third place if there’s a tie for first place.
- If the leader is forced to drop out of a race, the runner-up takes over this place.
- The square root is needed to calculate the length of the hypotenuse in this type of triangle.
- The graph of the declination of the Sun during a year can be approximated by this type of curve.

Answers

- six

- sum
- add
- nine
- FOIL
- half
- least
- mode
- two
- first
- right
- sine

### My 0.04 Seconds of Fame

In 2017, I attended the International KenKen Championship and filmed the final round, which I posted previously on this blog. But filmmakers Louis Cancel, Chris Flaherty, and Daniel Sullivan were there that day, too, and their cameras were significantly more sophisticated than my Samsung S8. Their footage of the competition, coupled with myriad interviews of competitors, organizers, and the inventor of KenKen himself, Tetsuya Miyamoto, has resulted in a new documentary, * Miyamoto and the Machine*, recently released by

*The New Yorker*. It tells the story of KenKen’s origins and attempts to answer the question, “Can a computer make puzzles as beautiful as those created by humans?”

Many aspects of the film will appeal to kenthusiasts, but my favorite moment occurs at 17:14. Competitor Ellie Grueskin is competing in the finals, and just over Ellie’s left shoulder is a barely visible, occasionally funny, middle-aged math guy holding — wait for it — a Samsung S8!

Yep, that’s me. I’m a star!

You have to ask yourself, what kind of monster would author such a shamelessly self-promotional post and not even provide one KenKen puzzle for the reader to enjoy? Definitely not me, so here you go.

After you solve the puzzle, definitely watch * Miyamoto and the Machine*. It’s 25 minutes well spent.

### More HIPE

Nearly five years ago, I wrote about HIPE, a parlor game in which one person gives a particular string of letters, and the other people in the parlor try to guess a word with that same string of letters (consecutively, and in the same order).

Well, I recently rediscovered *Can You Solve My Problems?* by Alex Bellos, and I was pleasantly surprised to find that he included four HIPEs in that book:

- ONIG
- HQ
- RAOR
- TANTAN

The fourth is one that I had included in my previous post, Don’t Believe the HIPE, and all are good enough that they deserve wide distribution.

Just for fun, here’s a new list of HIPEs that might prove interesting.

- SSP
- LWE
- NUSCU
- CUU
- CTW
- KGA
- UIU
- XII

In an effort to collect a bunch of excellent HIPEs, I’m asking for your help. If you play the game with friends and discover a particularly delectable combination of letters, please share below or at https://forms.gle/otddCw1uLeDALrMo7.

### A No-Op KenKen for Today

This will be a short post, just to share a puzzle for today.

There’s nothing inherently special about today — though it is the 30th anniversary of The Simpsons airing on Fox, and, slightly less important, the anniversary of Wilbur and Orville Wright’s famous flight — except that (a) I introduced the students in our middle school math club to KenKen last week, and (b) today is our last meeting before the holiday break, so I thought I’d do something special and create a KenKen puzzle that used the numbers from today’s date. I had hoped to include 12, 17, 20, and 19 as the target numbers in the cages, but that effort proved fruitless. Instead, I opted for 12, 1, 7, and 19 as the target numbers, and I filled in the single-cell cage in the bottom right with its number, 3.

I rather like the result. The puzzle is not terribly difficult; and, the solution is not unique, which I figure is perfect for kids who just learned about KenKen a week ago.

If you’re not familiar with No-Op KenKen, they’re just like regular KenKen puzzles, but the operation isn’t included with the target number. Instead, you’ll need to discern the operation for each cell. (For another example of a no-op KenKen puzzle, check out Harold Reiter’s No-Op 12 Puzzle.)

Enjoy, good luck, and happy December 17!

### 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 =

1512 – 3 =

912 × 3 =

3612 ÷ 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 4*n* + 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 ofx?

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!