Fun with hexadecimal

I recently placed the following question on an exam: “Convert 201,850,622 into base 16.” The answer: C07FEFE.

After returning the exams, I explained that there’s no V in base 16, so I had to settle for using 7 instead.

My Mathematical Magic Show: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The links below show the mathematical magic show that I’ll perform from time to time.

Part 1: Introduction.

Part 2a, Part 2b, and Part 2c: The 1089 trick.

Part 3a, Part 3b, and Part 3c: A geometric magic trick.

Part 4a: Part 4b, Part 4c, and Part 4d: A trick using binary numbers.

Part 5a, Part 5b, Part 5c, and Part 5d: A trick using the rule for checking if a number is a multiple of 9.

Part 7: The Fitch-Cheney card trick, which is perhaps the slickest mathematical card trick ever devised.

Part 8a, Part 8b, and Part 8c: A trick using Pascal’s triangle.

Part 9: Mentally computing $n$ given $n^5$ if $10 \le n \le 99$ .

Part 6: The Grand Finale.

And, for the sake of completeness, here’s a recent picture of me just before I performed an abbreviated version of this show for UNT’s Preview Day for high school students thinking about enrolling at my university.

English words in hexadecimal

Here’s a standard joke involving representing numbers in different bases.

Q: If only DEAD people understand hexadecimal, then how many people understand hexadecimal?

A: 57,005.

The joke, of course, is that $DEAD$ can be considered a number written in base 16, using the usual convention $A = 10$ , $B = 11$ , $C = 12$ , $D = 13$ , $E = 14$ , and $F = 15$ . In other words, $DEAD$ can be converted to decimal as follows:

$DEAD_{\small sixteen} = 13 \times 16^3 + 14 \times 16^2 + 10 \times 16 + 13 = 57,005$ .

After I heard this joke, I wondered just how many English words can be formed using only the letters A, B, C, D, E, and F so that I could make a subtle joke on a test. To increase the length of my list, I also allowed words that included the letters O (close enough to a 0), I (close enough to 1), and/or S (close enough to 5). However, I eliminated words that start with O (since a numeral normally doesn’t start with 0) and/or end in S (the plural version of these words are easily formed).

So I wrote a small program to search the dictionary that I have on my computer. The unabridged list follows, with words beginning with a capital letter (such as names or places) listed at the bottom. I emphasize that this list is unabridged, as there are several words on this list that I wouldn’t place on a test for obvious reasons: I would never ask my class to convert the base-10 numeral 721,077 into hexadecimal just so they can obtain the answer of $B00B5$ .

abaci

abase

abased

abbé

abed

abide

abided

abode

aboded

abscessed

abscissa

abscissae

acacia

accede

acceded

accessed

ace

aced

acid

acidic

acidified

add

added

ado

adobe

aid

aide

aided

aside

assessed

baa

baaed

babe

babied

bad

bade

baobab

base

based

basic

bassi

basso

bead

beaded

bed

bedded

bedside

bee

beef

beefed

beside

biased

biassed

bib

bid

bide

bided

boa

bob

bobbed

bode

boded

bodice

boo

boob

boobed

booed

bossed

cab

cabbed

cabbie

caboose

cacao

cad

caddied

café

cascade

cascaded

case

cased

cassia

cease

ceased

cede

ceded

cicada

cicadae

cob

cobbed

cocci

cocoa

cod

coda

codded

code

coded

codified

coed

coffee

coif

coifed

coiffed

coo

cooed

dab

dabbed

dad

dado

dead

deaf

deb

debase

debased

decade

decaf

decease

deceased

decide

decided

decode

decoded

deed

deeded

deface

defaced

defied

deice

deiced

deified

dice

diced

did

die

died

diocese

diode

disc

disco

discoed

disease

diseased

dissed

doc

dodo

doe

doff

doffed

doodad

dose

dosed

ease

eased

ebb

ebbed

eddied

edifice

edified

efface

effaced

facade

face

faced

fad

fade

faded

fed

fee

feed

fiasco

fib

fibbed

fie

fief

fife

fob

fobbed

foci

foe

food

ice

iced

idea

sac

sad

safe

said

sassed

scab

scabbed

scad

scoff

scoffed

sea

seabed

seafood

seaside

secede

seceded

see

seed

seeded

sic

side

sided

sob

sobbed

sod

soda

sodded

sofa

Abbasid

Abe

Acadia

Ada

Addie

Aida

Asia

Assad

Assisi

Basie

Bede

Beebe

Bessie

Bib

Bic

Bob

Bobbi

Bobbie

Boccaccio

Boise

Bose

Case

Casio

Cassie

Cid

Cobb

Dacca

Dada

Debbie

Dec

Decca

Dee

Defoe

Dido

Doe

Eco

Edda

Eddie

Effie

Essie

Feb

Fed

Fido

Iaccoca

Ibo

Ida

Isaac

Issac

Blue license plate

True story: while driving around town last month, I saw a bright blue car with the matching license plate of 0000FF.

Boy, I wish I had thought of this first.

My Mathematical Magic Show: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. Here’s my series on the mathematical magic show that I’ll perform from time to time.

Part 1: Introduction.

Part 2a, 2b, and 2c: The 1089 trick.

Part 3a, 3b, and 3c: A geometric magic trick (see also here).

Part 4a, 4b, 4c, and 4d: A trick using binary numbers.

Part 5a, 5b, 5c, 5d: Predicting a digit that’s been erased from a number.

Part 6: Finale.

Part 7: The Fitch-Cheney 5-card trick.

Part 8a, 8b, 8c: A trick using Pascal’s triangle.

My Mathematical Magic Show: Index

Part 1: Introduction.

Part 2a, Part 2b, and Part 2c: The 1089 trick.

Part 3a, Part 3b, and Part 3c: A geometric magic trick.

Part 4a: Part 4b, Part 4c, and Part 4d: A trick using binary numbers.

Part 5a, Part 5b, Part 5c, and Part 5d: A trick using the rule for checking if a number is a multiple of 9.

Part 7: The Fitch-Cheney card trick, which is perhaps the slickest mathematical card trick ever devised.

Part 8a, Part 8b, and Part 8c: A trick using Pascal’s triangle.

Part 6: The Grand Finale.

999 Megabytes

This made me laugh…

…though I would have laughed harder if the band was called 1023 Megabytes.

Source: https://www.facebook.com/1013TheBrew/photos/a.195619466488.123649.110527966488/10153674166916489/?type=3&theater

My Mathematical Magic Show: Part 5d

Last March, on Pi Day (March 14, 2015), I put together a mathematical magic show for the Pi Day festivities at our local library, compiling various tricks that I teach to our future secondary teachers. I was expecting an audience of junior-high and high school students but ended up with an audience of elementary school students (and their parents). Still, I thought that this might be of general interest, and so I’ll present these tricks as well as the explanations for these tricks in this series. From start to finish, this mathematical magic show took me about 50-55 minutes to complete. None of the tricks in this routine are original to me; I learned each of these tricks from somebody else.

Though this wasn’t part of my Pi Day magic show, I recently read an interesting variant on my fourth trick. The next time I do a mathematical show, I’ll do this trick next — not to amaze and stun my audience, but to see if my audience can figure out why it works. Each member of the audience will need to have a calculator (a basic four-function calculator will suffice). Here’s the patter:

I want you to take out your calculator. Using only the digits 1 through 9, there are three rows, three columns, and two diagonals. I want you to pick either a row, a column, or a diagonal. Then I want you to enter a three digit number using those numbers. For example, if you chose the first row, you can enter 123 or 312 or 231 or any three-digit number using each digit once.

Now, I want you to multiply this number by another three-digit number. So hit the times button.

(pause)

Now, choose another row, column, or diagonal and type in another three-digit number, using each of the three digits once.

(pause)

Now hit the equals button to multiply those two numbers together.

(pause)

Is everyone done? The product you just computed should have either five or six digits. I want you to concentrate on one of those digits. Just make sure that you concentrate on a digit other than zero, because zero is boring. So concentrate on a nonzero digit.

(pause)

(I point to someone.) Without telling me the digit you chose, please tell me the other digits in your product.

The audience member will say something like, “3, 7, 9, and 2.” To which I’ll reply in three seconds or less, “The number you chose was 6.”

Then I’ll turn to someone else and ask which numbers were not scratched out. She’ll say something like, “1, 1, 9, 7, and 2.” I’ll answer, “The number you chose was 7.”

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

Clearly this works using the same logic as my fourth magic trick: the product is always a multiple of 9, and so I can add the digits to figure out the missing digit. The more interesting question is: Why is the product always a multiple of 9?

This works because each of the factors of the product is a multiple of 3. Let’s take another look at the calculator.

If the first row is chosen, the sum of the digits is 1+2+3 = 6, a multiple of 3. And it doesn’t matter if the number is 123 or 312 or 231… the order of the digits is unimportant.

If the second row is chosen, the sum of the digits is 4+5+6 = 15, a multiple of 3.

If the third row is chosen, the sum of the digits is 7+8+9 = 24, a multiple of 3.

If the first column is chosen the sum of the digits is 1+4+7=12, a multiple of 3.

If the second column is chosen, the sum of the digits is 2+5+8 = 15, a multiple of 3.

If the third column is chosen, the sum of the digits is 3+6+9 = 18, a multiple of 3.

If one diagonal is chosen, the sum of the digits is 1+5+9 = 15, a multiple of 3.

If the other diagonal is chosen, the sum of the digits is 3+5+7 = 15, a multiple of 3.

This can be stated more succinctly using algebra. The digits in each row, column, and diagonal form an arithmetic sequence. For each row, the common difference is 1. For each column, the common difference is 3. And for a diagonal, the common difference is either 2 or 4. If I let $a$ be the first term in the sequence and let $d$ be the common difference, then the three digits are $a$ , $a + d$ , and $a + 2d$ , and their sum is

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

which is a multiple of 3. (Indeed, the sum is 3 times the middle number.)

So each factor is a multiple of 3. That means the product has to be a multiple of $9$ . In other words, if the first factor is $3m$ and the second factor is $3n$ , where $m$ and $n$ are integers, their product is equal to

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

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

My Mathematical Magic Show: Part 5c

I want you to take out your calculator. Using only the digits 1 through 9, there are three rows, three columns, and two diagonals. I want you to pick either a row, a column, or a diagonal. Then I want you to enter a three digit number using those numbers. For example, if you chose the first row, you can enter 123 or 312 or 231 or any three-digit number using each digit once.

Now, I want you to multiply this number by another three-digit number. So hit the times button.

(pause)

Now, choose another row, column, or diagonal and type in another three-digit number, using each of the three digits once.

(pause)

Now hit the equals button to multiply those two numbers together.

(pause)

Is everyone done? The product you just computed should have either five or six digits. I want you to concentrate on one of those digits. Just make sure that you concentrate on a digit other than zero, because zero is boring. So concentrate on a nonzero digit.

(pause)

(I point to someone.) Without telling me the digit you chose, please tell me the other digits in your product.

The audience member will say something like, “3, 7, 9, and 2.” To which I’ll reply in three seconds or less, “The number you chose was 6.”

Then I’ll turn to someone else and ask which numbers were not scratched out. She’ll say something like, “1, 1, 9, 7, and 2.” I’ll answer, “The number you chose was 7.”

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

Clearly this works using the same logic as yesterday’s post: the product is always a multiple of 9, and so I can add the digits to figure out the missing digit. The more interesting question is: Why is the product always a multiple of 9? I’ll address this in tomorrow’s post.

My Mathematical Magic Show: Part 5b

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

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

(pause)

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

(pause)

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

(pause)

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

(pause)

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

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

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

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

$43,125-24,513$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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