## Posts tagged ‘trick’

### 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!

### 5 People Born in the 1800’s — and Still Alive!

There’s been a big deal made this week about the five people still alive who were born in the 1800’s, and rightfully so. As I wash down an A1 Peppercorn Burger from Red Robin and a bag of Cheetos with a Dr Pepper, I’m not even sure I’ll make it to 50, let alone 115.

I feel bad for these five women. The following internet math trick claims that it works “for everyone in the whole world,” but it doesn’t work for supercentenarians:

- Take the last two digits of the year in which you were born.
- Now add the age you will be this year.
*The result will be 115 for everyone in the whole world.*

For instance, the oldest living person, Misao Okawa, was born on March 5, 1898. For her, this trick yields 98 + 117 = 215, not 115 as promised.

As it turns out, this trick doesn’t work for anyone born after 2000, either. For instance, my sons were born May 2, 2007, and for them, 07 + 8 = 15, not 115.

Now, I know what you’re thinking. Surely, something must be wrong. There’s not really an error floating around the internet, right? But it does appear to be the case.

Luckily, I have a solution for this problem. My modification of the trick is as follows, and then it really will work for everyone in the whole world:

- Take the year in which you were born.
- Now add the age you will be this year.
*The result will be 2015 for everyone in the whole world.*

Unsatisfying, sure, but at least it’s correct.

There are lots of reasons to which the very old attribute their longevity. Chief among them is having never taken a statistics course at a community college. But not far behind are eating healthy, exercising regularly, remaining active, having friends, and staying happy. Some simply attribute it to “living right.”

A reporter asked a centenarian, “To what do you attribute your longevity?”

“Simple,” said the man. “I never argue.”

“Oh, surely there must be more to it than that,” said the reporter.

“Well,” said the elderly man. “I guess you’re right.”

It’s not uncommon for reporters to interview centenarians and ask them about their longevity.

A 100-year old man was asked, “To what do you attribute your long life?”

“I’m not quite sure yet,” he replied. “I’m still negotiating with two cereal companies.”

And finally, to celebrate her 100th birthday, Dorothy Carchman was invited on-stage with the cast of *Old Jews Telling Jokes*. She delivered the following punch line with aplomb:

Becky, who was 90 years old, and her best friend, Dorothy, are driving down the road. Becky said, “Dorothy, that’s the third car you almost hit in five minutes.”

And Dorothy replied, “Wait… I’m driving?”

### Can I Get Your Digits?

Saw this on a t-shirt recently:

It made me think about this problem involving digits.

Consider the number

1234567891011121314151617181920obtained by writing the numbers from 1 to 20 in order side-by-side.

What’s the greatest number that can be obtained by crossing out 20 digits?

If a fetching lady or handsome gent catches your fancy by solving that problem, you might want to ask her or him…

How can I know so many digits of π and so few digits of your phone number?

And if he or she still hasn’t taken leave of you, then you could really press your luck with the following:

- Ask your new friend to write down a number with four or more digits.

459,163

- Then, have your friend add the digits.

4 + 5 + 9 + 1 + 6 + 3 = 28

- Subtract the sum from the original number.

459,163 – 28 = 459,135

- Have your friend cross out one of the digits, and then read the remaining number aloud to you.

45,935

- Then, miraculously announce the missing digit.

**1**

The secret to the trick? Simple. Just add the digits of the number that your friend reads aloud, and then figure out what number must be added to get the sum to a multiple of 9. Above, the digits of the number 45,935 have a sum of 26, which is 1 less than a multiple of 9, so the removed digit is 1.

### Every (Math) Trick in the Book

Linda Gojak has long been a proponent of doing away with tricks in math class (see also *Making Mathematical Connections*, October 2013).

She has plenty of company from Tina Cardone, author of Nix the Tricks, a free downloadable book of tricks that Cardone believes should be removed from the curriculum. (It also includes what and how things should be taught instead. Though Cardone is the author, she admits it was a collaborative effort of #MTBoS.)

But Gojak and Cardone are not talking about fun and mysterious math tricks, like this:

- In a bookstore or library, pick out two identical copies of the same book. Keep one, and give one to your friend.
- Ask your friend to pick a number, and both of you turn to that page in the book.
- Then tell your friend to pick a word in the top half of that page, and silently spell that word while moving forward one word for each letter. (For instance, if they picked “math,” they’d move forward four words as they spelled.) Have them do that again with the word they land on; then again; and again; till they reach the end of the page. They should stop at the last word that won’t take them to the next page.
- Do the same thing with your copy of the book — though make it obvious that you’re not copying them and using the same first word.
- Miraculously, you and your friend will land on the same word at the end of the page.

What Gojak and Cardone are talking about and, rightfully, railing against are the mnemonics and shortcuts that many teachers give to students in lieu of developing true understanding, such as:

*Ours is not to reason why; just invert, and multiply.*- Just add a 0 when multiplying by 10.
- The butterfly method of adding and subtracting fractions.

From my perspective, tricks are not inherently bad, as long as students have developed the conceptual understanding necessary to recognize why the trick works. In fact, the standard algorithms for multiplication and long division are, essentially, math tricks. But if students are able to reason their way to an answer — and, even better, if they can discern these tricks on their own from examples — then the ‘tricks’ can be useful in helping them get answers quickly.

**The problem is that many teachers (or well-meaning parents) introduce tricks before students understand the underlying mathematics**, and that can be to the students’ detriment. Instead of mathematics appearing to be a cohesive whole, students believe that it is a large bag of disconnected procedures to be memorized, remembered for the test, and then promptly forgotten.

My eighth-grade English teacher taught us the following:

You need to learn the rules of grammar, so you can feel comfortable breaking them.

I believe the corollary in mathematics applies here. If you understand the structure and algorithms associated with procedural fluency, then you can then feel comfortable using well-known shortcuts.

The Common Core identifies three pillars of rigor — **conceptual understanding**, **procedural fluency**, and **application** — and asserts that each pillar must be “pursued with equal intensity” (CCSSM Publishers’ Criteria for K-8 and High School). But I’ll take that one step further. I contend that conceptual understanding must happen **prior** to procedural fluency for learning to be effective. That contention is based on research as well as personal experience, and it’s consistent with the NCTM *Procedural Fluency in Mathematics *Position Statement, in which the Council states, “Procedural fluency builds on a foundation of conceptual understanding, strategic reasoning, and problem solving,” citing the *Common Core State Standards for Mathematics*, *Principles and Standards for School Mathematics*, and* Principles to Actions: Ensuring Mathematical Success for All.*

A mathematical trick, as far as I’m concerned, is merely the recognition and application of a pattern. A simple example is the 9× facts.

- 1 × 9 = 9
- 2 × 9 = 18
- 3 × 9 = 27
- 4 × 9 = 36
- etc.

By looking for patterns, students might notice that 9 times a number is equal to 10 times that number minus the number itself; that is, 9*n* = 10*n* – *n*. For instance, 8 × 9 = 80 – 8. This makes sense algebraically, of course, since 10*n* – *n* = (10 – 1)*n* = 9*n*, but I wouldn’t expect a second grader to be able to work this out symbolically. I might, however, expect a second grader to correctly reason that this should always work. From that, a student would have a way to remember the 9× facts that is based on experience and understanding, not just remembering a rule that was delivered from on high.

Taking this further, recognizing this pattern helps with multiplication problems beyond the 9× facts. Recognizing that 9*n* = 10*n* – *n* can also help a student with the following exercise:

- 9 × 26 = 260 – 26 = 234

As a more sophisticated example, students might be able to generalize a rule for multiplying two numbers in the teens. I’ve seen tricks like this presented as follows, with no explanation:

- Take two numbers from 11 to 19 whose product you’d like to know.
- For example, 14 × 17.

- Add the larger number to the units digit of the smaller number.
- 17 + 4 = 21

- Concatenate a 0 to the result.
- 21(0) = 210

- Multiply the units digits of the two numbers.
- 7 × 4 = 28

- Add the last two steps.
- 210 + 28 = 238

- And there you have it: 17 × 14 = 238.

Why this trick works is rather beautiful.

17 × 14

(10 + 7) × (10 + 4)

10 × 10 + 7 × 10 + 10 × 4 + 7 × 4

100 + 70 + 40 + 28

**10 **×** (17 + 4)** + 28

210 + 28

238

The steps requiring students to *add the larger number to the units digit of the smaller number* and then *concatenate a 0 to the result* are represented in bold above as **10 × (17 + 4)**. Those steps in the rule are just a shortening of the second, third, and fourth lines in the expansion above.

Presenting this trick with an explanation is better than presenting it without explanation, for sure, but better still would be for students to look for patterns from numerous examples and, if possible, generate their own rules. If you presented students with the following list:

- 15 × 16 =240
- 17 × 14 = 238
- 13 × 18 = 234
- 19 × 12 = 228

would they be able to generate the above rule on their own? Probably not. But that’s okay. The rule is not the point. The point is that there is an inherent structure and pattern within numbers, and students might notice some patterns in the factors and products in the four examples above — for instance, the smaller factor decreases from 15 to 14 to 13 to 12 as you progress from one example to the next; the larger factor increases from 16 to 17 to 18 to 19; and the product decreases by 2, then by 4, then by 6.

Something’s going on there, but what?

*Ours is not to tell them why; just give them more examples to try!*

### A Date of Good Luck

Today is 11/7/13, which is a semi-lucky/lucky/unlucky date.

The numbers in today’s date are 7, 11, and 13, which are the same numbers used in my favorite math trick. I shared that trick on 7/11/13 in A Great Day for a Math Trick — which, honestly, was the best day ever to share it (unless you’re in Europe, in which case today is perfect). For U.S. audiences today, do the division by 11 before the division by 7:

- Multiply your age by
**12**. - Now
**add the age**of your spouse/brother/sister/friend/uncle/aunt/whomever. - This should yield a three-digit number. Now, divide by
**11**. - Then, divide by
**7**. - Then, divide by
**13**. - The result should be a number of the form 0.
*abcdef*…, with a 0 and a decimal point in front of a long string of digits. Add the**first six digits**after the decimal point.

Here’s the cool part. I don’t know your age, nor do I know the age of your spouse, brother, sister, friend, uncle, or aunt. But I do know that after you completed those steps, **this is your result**.

Now, how did I know that?

That’s for you to figure out.

Math Major: I’ve found that 67% of Literature majors are stupid.

Literature Major: I’m part of the other 13%.I was walking past a mental hospital the other day, and all the patients were shouting, “13… 13… 13… 13.” The fence was very high, so I peeked through a little gap in the planks to see what was going on.

Some bastard poked me in the eye with a stick.

Then they all started shouting, “14… 14… 14… 14.”

### Another Bad Email Math Puzzle…

It happened again. I received another email with a number trick that makes the ubiquitous claim, “This will work for everyone!” Sadly, it won’t, but it was kind of cool:

Calculate 39 × (your age) × 259.

The email said, “The result will surprise you.”

It didn’t. I suspected what the value of 39 × 259 would be, so I predicted the result. But if you don’t know the value of that product, then maybe you’ll be surprised.

The trick works well enough if you have a double-digit age. But my friend Ferdinand is 107 years old. His result was 1,080,807, and that just looks like a mess. The results for my six-year-old sons were better, albeit rather unsatisfying.

Ha-rumph. So much for the Internet providing mathematical inspiration.

Here are some similarly uninteresting puzzles that I created:

Calculate 7,373 × (your age) × 137.

Calculate 9,091 × (your age) × 11,111.

Calculate 101 × (your age) × 1,000,100,010,001.

To create more puzzles like this, enter **factor(101010…10101)** into Wolfram Alpha.

For your centenarian friends, try these:

Calculate 101,101 × (your age) × 9,901.

Calculate 3.3 × (your age) × 33.67.

Let’s not forget the little people whose age is still in the single digits:

Calculate 3 × (your age) × 37.

Calculate 41 × (your age) × 271.

And a math joke (or is it?) about age:

I’ve been good with numbers my whole life. When I turned 2, I realized that my age had doubled in one year. This concerned me… at that rate, I’d be 32 in four more years!

And another:

What goes up but never comes down?

Your age.

### A Great Day for a Math Trick

Today is 7/11/13, and boy, have I got a great math trick for today! You’ll likely need a calculator.

- Multiply your age by 12.
- Now add the age of your spouse/brother/sister/friend/uncle/aunt/whomever.
- This should yield a three-digit number. Now, divide by 7.
- Then, divide by 11.
- Then, divide by 13.
- The result should be a number of the form 0.
*abcdef*…, with a 0 and a decimal point in front of a long string of digits. Add the**first six digits**after the decimal point.

Here’s the cool part. I don’t know your age, nor do I know the age of your spouse, brother, sister, friend, uncle, or aunt. But I do know that after you completed those steps, the result was 27.

Pretty cool, eh?

There are myriad math tricks of this ilk, but this one is my favorite. It’s based on a trick I learned from Art Benjamin, though I think the one above has more panache than his original. Decide for yourself.

- Choose a number from 1 to 70, and then divide it by 7.
- If your total is a whole number (that is, no digits after the decimal point), divide the answer by 7 again.
- Is there a 1 somewhere after the decimal point? I predict that the number after the 1 is 4. Am I right?
- Now add up the first six digits after the decimal point.

Just as with the trick above, the result will always be 27.

Regardless of which trick you prefer, have a happy 7/11! And if you’ve got a few hours to kill, you can try to solve the 7‑11 problem.

### Is It Yours? It’s Not Mayan…

It’s December 21. You’re here. I’m here. So much for the prophecy of the Mayan calendar.

So, will someone please call Ms. Angelou and tell her she had it wrong?

Actually, the Mayan calendar never predicted the apocalypse. (Nor was it developed by Maya Angelou. Or Maya Rudolph. or Maya Lin. Or anyone else named Maya.) In truth, one cycle of the Mayan calendar is ending, so a new cycle is about to begin. It’s not a like a time bomb that will explode when the cycle ends. It’s more like the odometer of a car rolling over.

While I can forgive folks who misread the Mayan calendar, I have less patience for folks who misunderstand *our* calendar.

I recently received an email that stated the following:

This year, December has five Saturdays, five Sundays, and five Mondays. This will only happen once every 824 years.

Oish. Really? I wish that folks who forward this kind of nonsense would, at a minimum, look at a calendar. (At a maximum, I wish they would lose my contact info.)

The good folks at www.timeanddate.com will gladly show you the calendar for December of any year you like. And if you look at the calendar for December in 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, 2096, or any of 105 other years within 824 years of today, you’ll see that they all have five Saturdays, five Sundays, and five Mondays. Consequently, it doesn’t seem that December 2012 is terribly special.

The folks at www.timeanddate.com also have a nice explanation of why the math in the email that I received is all wrong (though their article is based on July 2011 which had five Fridays, five Saturdays and five Sundays, and they disprove an argument saying that such an occurrence happens once every 823 years; but, whatever).

I suspect that most folks are unaware that our calendar repeats in a 28‑year cycle. And I’d bet that even fewer realize there is a nice pattern of 6‑11‑6‑5 years when the calendar repeats… assuming you skip those nasty century years, like 1900 and 2100, that fail to include a leap day.

Still, I think most reasonably intelligent humans should recognize that a claim like “only once every 824 years” has to be an exaggeration.

But perhaps that’s the problem: I’m assuming that people who forward emails like this are reasonably intelligent.

Along similar lines, here’s a math trick that I’ve received several times via email:

- Take the last two digits of the year in which you were born.
- Now add the age you will be this year. (That is, if you’ve already had your birthday this year, add your current age. If you haven’t, add the age you’ll turn on your birthday this year.)
- The result will be 112 for
*everyone in the whole, wide world*.

There’s only one problem with this trick: It doesn’t work.

For someone like Besse Cooper, who was born in 1896, the result will be 212.

For someone like my twin five-year-old sons, who were born in 2007, the result will be 12.

In fact, the trick won’t work for anyone born before 1900 or after 2000. Based on data about age distribution, the result will not be 112 for approximately 15% of the U.S. population. The yellow bars in the graph below indicate the ages for which this trick does not work.

A better statement of this “trick” might be…

- Take the year in which you were born.
- Now add the age that you will be this year.
- The result will be 2012 for everyone in the whole, wide world.

Completely correct! But not much of a trick anymore, is it?

### In Your Prime

I’ve got a prime number trick for you today.

- Choose any prime number
*p*> 3. - Square it.
- Add 5.
- Divide by 8.

Having no idea which prime number you chose, I can tell you this:

The remainder of your result is 6.

Pretty cool, huh?

I will now fill a bunch of space with quotes and jokes about prime numbers to prevent you from seeing the spoiler explanation below. But you can skip straight to the bottom if you’re not interested in the other stuff or if you just can’t control yourself.

Mark Haddon, author of *The Curious Incident of the Dog in the Night-time*, wrote the following:

Prime numbers are what is left when you have taken all the patterns away. I think prime numbers are like life. They are very logical, but you could never work out the rules, even if you spent all your time thinking about them.

(Incidentally, if you haven’t read that book, you should. Amazon reviewer Grant Cairns said it better than I could: “The integration of the mathematics into the fiction is better than any other work that I know of. The overall effect is a beautiful story that any maths fans will find hard to read without the tissue box close at hand.”)

Israeli mathematician Noga Alon said that he was interviewed on Israeli radio, and he mentioned that Euclid proved over 2,000 years ago that there are infinitely many primes. As the story goes, the host immediately interupted him and asked:

Are there still infinitely many primes?

And of course there’s this moldy oldie:

Several professionals were asked how many odd integers greater than 2 are prime. The responses were as follows:

Mathematician:3 is prime, 5 is prime, 7 is prime, and by induction, every odd integer greater than 2 is prime.

Physicist:3 is prime, 5 is prime, 7 is prime, 9 is experimental error, 11 is prime, …

Engineer:3 is prime, 5 is prime, 7 is prime, 9 is prime, 11 is prime, …

Programmer:3 is prime, 5 is prime, 7 is prime, 7 is prime, 7 is prime, …

Marketer:3 is prime, 5 is prime, 7 is prime, 9 is a feature, …

Software Salesperson:3 is prime, 5 is prime, 7 is prime, 9 will be prime in the next release, …

Biologist:3 is prime, 5 is prime, 7 is prime, the results for 9 have not yet arrived…

Advertiser:3 is prime, 5 is prime, 7 is prime, 11 is prime, …

Lawyer:3 is prime, 5 is prime, 7 is prime, there is not enough evidence to prove that 9 isnotprime, …

Accountant:3 is prime, 5 is prime, 7 is prime, 9 is prime if you deduct 2/3 in taxes, …

Statistician:Try several randomly chosen odd numbers: 17 is prime, 23 is prime, 11 is prime, …

Professor:3 is prime, 5 is prime, 7 is prime, and the rest are left as exercises for the student.

Psychologist:3 is prime, 5 is prime, 7 is prime, 9 is prime but tries to suppress it, …

Card Counter:3, 5, and 7 are all prime, but I prefer 21.

*Explanation of the Prime Number Trick*

We are trying to show that (*p*^{2} + 5) mod 8 = 6. This is equivalent to showing that (*p*^{2} ‑ 1) mod 8 = 0, or that (*p* + 1)(*p* ‑ 1) is divisible by 8.

Because *p* > 3 and is prime, then either *p* = 1 mod 4 or *p* = 3 mod 4. Consequently, it must be the case that (a) *p* + 1 = 2 mod 4 and p ‑ 1 = 0 mod 4 or (b) *p* ‑ 1 = 2 mod 4 and *p* + 1 = 0 mod 4. That is, both numbers will be even, and at least one of them will be a multiple of 4. For either (a) or (b), the product (*p* + 1)(*p* ‑ 1) will be a multiple of 8. Q.E.D.

### Rank Math Trick

The Amazon sales rank for *Math Jokes 4 Mathy Folks* fluctuates between 20,000 and 300,000. The highest rank to date was 11,394, which occurred July 7, 2010, the morning after a review of the book appeared in Maria Miller’s Homeschool Math Blog and in her email newsletter.

Today, the Amazon sales rank was 113,113. Cool number, eh? Reminds me of an arithmetrick:

- Take a three‑digit number
*abc*. Then, write it twice to make a six‑digit number*abc*,*abc*. (For instance, if you chose 113, then your six‑digit number would be 113,113.) - I’m feeling lucky, so divide by 7.
- Hmm… I’m not feeling quite as lucky now, so divide by 11.
- Uh-oh. I no longer feel lucky at all. Divide by 13.
- Check your result. Should be the three‑digit number you started with,
*abc*.

The following joke is a hint to why this trick works, in case you haven’t already figured it out:

Teacher: Can you find the prime factorization of 1,001?

Student: I didn’t even know it was lost!