One Equals Zero! The following is a "proof" that one equals zero.
Consider two non-zero numbers x and y such that
x = y.
Then x2 = xy.
Subtract the same thing from both sides:
x2 - y2 = xy - y2.
Dividing by (x-y), obtain
x + y = y.
Since x = y, we see that
2 y = y.
Thus 2 = 1, since we started with y nonzero.
Subtracting 1 from both sides,
1 = 0.
What's wrong with this "proof"?
Presentation Suggestions:
This Fun Fact is a reminder for students to always check when they are dividing by unknown variables for cases where the denominator might be zero.
The Math Behind the Fact:
The problem with this "proof" is that if x=y, then x-y=0. Notice that halfway through our "proof" we divided by (x-y).
Memorizing Pi
The digits of Pi are fascinating. As the ratio of the circumference of a circle to its diameter, Pi has such a fundamental definition, and yet this ratio is irrational and so its decimal expansion never repeats. It is easy to be mesmerized by the digits of the decimal expansion:
3.14159265358979323846264338327950288419716939937510...
and many people have tried to memorize digits of pi for fun.
One fun way to memorize the first few digits is to use sentence mnemonics for pi--- phrases in which the number of letters of each successive word corresponds to a digit of pi. Here are some well-known pi mnemonics:
"Wow! I made a great discovery!" (3.14159...)
"Can I have a small container of coffee?" (3.1415926...)
"How I want a drink, alcoholic of course, after the heavy lectures involving quantum mechanics." (3.14159265358979...)
Kaprekar's Constant
Take any four digit number (whose digits are not all identical), and do the following:
Rearrange the string of digits to form the largest and smallest 4-digit numbers possible.
Take these two numbers and subtract the smaller number from the larger.
Use the number you obtain and repeat the above process.
What happens if you repeat the above process over and over? Let's see...
Suppose we choose the number 3141.
4311-1134=3177.
7731-1377=6354.
6543-3456=3087.
8730-0378=8352.
8532-2358=6174.
7641-1467=6174...
The process eventually hits 6174 and then stays there!
But the more amazing thing is this: every four digit number whose digits are not all the same will eventually hit 6174, in at most 7 steps, and then stay there!
Presentation Suggestions:
Remember that if you encounter any numbers with fewer than has fewer 4 digits, it must be treated as though it had 4 digits, using leading zeroes. Example: if you start with 3222 and subtract 2333, then the difference is 0999. The next step would then consider the difference 9990-0999=8991, and so on. You might ask students to investigate what happens for strings of other lengths or in other bases.
The Math Behind the Fact:
Each number in the sequence uniquely determines the next number in the sequence. Since there are only finitely many possibilities, eventually the sequence must return to a number it hit before, leading to a cycle. So any starting number will give a sequence that eventually cycles. There can many cycles; however, for length 4 strings in base 10, there happens to be 1 non-trivial cycle, and it has length 1 (involving the number 6174).
Mind-Reading Number Trick
Think of a number, any positive integer (but keep it small so you can do computations in your head).
1. Square it.
2. Add the result to your original number.
3. Divide by your original number.
4. Add, oh I don't know, say 17.
5. Subtract your original number.
6. Divide by 6.
The number you are thinking of now is 3!
How did I do this?
Presentation Suggestions:
Ham it up with magician's patter. Step 4 could be anything you want---someone's age, or their favorite number--- just ask the crowd for suggestions. (This will change the final outcome of Step 6, but see below for how.)
The Math Behind the Fact:
Clearly no matter what you start with, the answer should come out the same (zero wasn't allowed because of Step 3). We can see why this trick works by using a little bit of high school algebra! If you follow the instructions starting with the variable X instead of an actual number, you will see that X is eliminated by Step 5.
Using this idea, you can make up your own mental math trick right on the spot! (Just don't do anything too obvious, like tell people to add 5, subtract their original number, and say "the number you are thinking of is 5".)
Magic 1089
Here's a cool mathematical magic trick. Write down a three-digit number whose digits are decreasing. Then reverse the digits to create a new number, and subtract this number from the original number. With the resulting number, add it to the reverse of itself. The number you will get is 1089!
For example, if you start with 532 (three digits, decreasing order), then the reverse is 235. Subtract 532-235 to get 297. Now add 297 and its reverse 792, and you will get 1089!
Presentation Suggestions:
You might ask your students to see if they can explain this magic trick using a little algebra.
The Math Behind the Fact:
If we let a, b, c denote the three digits of the original number, then the three-digit number is 100a+10b+c. The reverse is 100c+10b+a. Subtract: (100a+10b+c)-(100c+10b+a) to get 99(a-c). Since the digits were decreasing, (a-c) is at least 2 and no greater than 8, so the result must be one of 198, 297, 396, 495, 594, 693, 792, or 891. When you add any one of those numbers to the reverse of itself, you get 1089!
Red-Black Card Trick
Here's a pretty easy card trick that you can do that can also be pretty surprising. Here's how the trick you do will appear to others:
Take a deck of cards, and give it to a spectator and ask her to shuffle the deck and return it to you face down. You take the cards, and (with a little showmanship but without looking at the fronts of the cards) separate them into two piles, and then say "Just by feeling the redness or blackness of the cards with my fingers, I've made two piles so that the number of red cards in the first pile is the number of black cards in the second pile."
Have your spectator turn over the cards and verify!
Presentation Suggestions:
Your spectator can shuffle the cards as many times as she likes--- it won't matter! When she gives the cards to you, all you are really doing (though don't make it obvious) is counting the cards into two piles so that there are 26 cards in each pile.
The Math Behind the Fact:
The reason this trick works is simple... if the number of red cards in the first and second piles is R and S, and the number of black cards in the first and second piles is A and B, then we know that R+S=26 (since the total number of red cards is 26) and S+B=26 (since the total number of cards in the second pile is 26). These two equations can be subtracted from one another to show that R-B=0, or R=B.
Red-Black Pairs Card Trick
Here's a terrific mathematical card trick that will impress your friends. When you do this trick, the effect of the card trick will look like this:
You have a deck of cards, and you ask for a volunteer who knows how to do a riffle shuffle. You then cut the deck and then give the volunteer the halves of the deck and ask him to do one riffle shuffle and return the deck to you. Now say "There's no way I could know anything about the deck right now, right? Well, I was born with the amazing ability to feel the redness and blackness of cards with my fingertips. However, my talent is not that refined. I can only feel red and black cards in pairs." As you say this, put the deck of cards behind your back (so that you cannot see them) and then, at regular intervals, you fish around in the deck and pull out pairs of cards and show them to the audience. These pairs will all have exactly one black and one red card!
Presentation Suggestions:
Before performing the trick, order the deck alternating colors, all the way through, red-black-red-black-... etc. (When you flash the deck before their eyes, they really won't notice this pattern if you do it quickly.)
After this, there is really only one thing you need to remember to ensure that the trick works: you must cut the deck (not the spectator), and you must do it in such a way that the bottom of each half of the deck is a different color. Then, no matter how the spectator riffle shuffles the deck, the cards will always drop in red-black or black-red pairs. See below for explanation.
Then, all you have to do after the deck is returned and you put it behind your back is to pull out the top 2 cards. It will be either red-black or black-red! Then pull out the next 2 cards, which again will be red-black or black-red. You can continue in this fashion to the end of the deck, if you like!
Of course, you should make it look as if you are trying really hard to find the cards (even though what you are really doing is very easy). Spectators will wonder if you are pulling one card off the top and one card off the bottom; but you can pull the deck out and show them that this is not the case.
The Math Behind the Fact:
The reason the trick works at the point of the riffle shuffle is both simple and stunning: if you cut the deck so that the cards at the bottom of each half are different colors, then the first card that gets "dropped" in the shuffle will be a different color then the second card that gets dropped, no matter which half of the deck they come from. As an example, if the first card that gets dropped is black, then after that both halves will have red cards at the bottom, so no matter which card falls next it will be red! After this, both halves again have different colored cards at bottom and we are back to the situation at start. So all the cards will fall off in either red-black or black-red pairs.
The message of this trick is that one shuffle is not enough to randomize a deck of cards-- you really can know something about the deck after one shuffle... but only if you stack the deck in a particular way first!
Binary Card Trick
You put a deck of cards in your pocket, and invite anyone in the audience to call out a number between 1 and 15. Then you reach into your pocket, you take out a set of cards whose sum is the number that was called!
How can you perform this magic trick?
The Math Behind the Fact:
This mathematical magic trick can be found in the reference and is based on the properties of binary numbers. Every number between 1 and 15 has a unique representation as a sum of some collection of the numbers 1, 2, 4, and 8. (To see which collection, just take the given number and successively subtract the largest number of 1, 2, 4, and 8 that is less than the given number. That number is part of your collection. The subtraction yields a new number; now repeat the process with this number, over and over, until you get 0.) The collection of numbers you obtain reveals the binary decomposition of the given number into sums of powers of two (in contrast to the usual representation of a number into sums of powers of ten).
Now before the trick starts, pick an Ace, 2, 4, and 8 and put them on top of the deck, and then put the deck in your pocket.
Then when a number between 1 and 15 is called out, take the binary decomposition of the number, and use that to determine which of the first four cards you will pull out. No one needs to know that you never need to use the other cards!
Leapfrog Addition
Here's a nice mathematical magic trick based on properties of the Fibonacci sequence.
Give your friend a card with ten blank lines, numbered 1 through 10. Have your friend think of two numbers between 1 and 20 and write them down on the first 2 lines of the card. Now in each of the successive lines, have your friend write the sum of the previous two lines. For instance, in line 3, write the sum of lines 1 and 2. In line 4, write the sum of lines 2 and 3, etc. until finally in line 10, your friend has written the sum of lines 8 and 9.
Ask your friend to total the numbers. If you've practiced the Multiplication by 11 Fun Fact, you'll be able to tell your friend the total faster than she can add the numbers (because the total will be just 11 times the number in line 7). Also, you can announce the quotient of line 10 divided by line 9... to 2 decimal places, it will be 1.61!
Let's do an example. Suppose your friend tells gives you the numbers 3 and 7. Her card will then have these numbers:
7
3
10
13
23
36
59
95
154
249
The total is 649 (which is just 11 times 59, do this in your head with the Multiplication by 11 Fun Fact.
The quotient 249/154 is 1.61 (to 2 decimal places).
Presentation Suggestions:
Write down the quotient 1.61 on another card, and place it in an envelope before the start of your magic trick. Then you can have your friend open that envelope after she has computed the quotient.
The Math Behind the Fact:
The trick works for the following reason. If the number X is in line 1, and the number Y is in line 2, then the number X+Y will be in line 3, the number (X+Y)+Y=(X+2Y) will be in line 4, and so on. Continuing, you will find that line 7 contains (5X+8Y), line 9 contains (21X+34Y), and line 10 contains (55X+88Y), which is indeed just 11 times line 7.
For the ratio of line 10 divided by line 9, we appeal to a property of "adding fractions badly": for positive numbers A, B, C, D where (A/B) < (C/D), it is a neat fact that the fraction you get by "adding badly": (A+C)/(B+D), must lie in between the values (A/B) and (C/D). So the ratio (55X+88Y)/(21X+34Y) must lie in between (55X/21X)=1.615... and (88Y/34Y)=1.619...
An even more stunning fact is that if you continue this leapfrog procedure with many more lines, then the ratio of successive lines will approach the golden ratio: 1.6180339... (If you know some linear algebra, this follows because the leapfrog procedure can be written as a 2-dimensional linear system of equations, and the largest eigenvalue of this system happens to be the golden ratio.)
This magic trick may be found in the delightful book in the reference.
Squares Ending in 5
Give me any 2 digit number that ends in 5, and I'll square it in my head!
452 = 2025
852 = 7225, etc.
There's a quick way to do this: if the first digit is N and the second digit is 5, then the last 2 digits of the answer will be 25, and the preceding digits will be N*(N+1).
Presentation Suggestions:
After telling the trick, have students see how fast they can square such numbers in their head, but doing several examples.
The Math Behind the Fact:
You may wish to assign the proof as a fun homework exercise: multiply (10N+5)(10N+5) and interpret! The trick works for larger numbers, too, although it may be harder to do this in your head. For instance 2052 = 42025, since 20*21=420. Also, you can combine this trick with other lightning arithmetic tricks. So 1152 = 13225, because 11*12 = 132, using the Multiplication by 11 trick.
The reference also contains more secrets of fast mental calculations.
Fibonacci Number Formula
The Fibonacci numbers are generated by setting F0=0, F1=1, and then using the recursive formula
Fn = Fn-1 + Fn-2
to get the rest. Thus the sequence begins: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... This sequence of Fibonacci numbers arises all over mathematics and also in nature.
However, if I wanted the 100th term of this sequence, it would take lots of intermediate calculations with the recursive formula to get a result. Is there an easier way?
Yes, there is an exact formula for the n-th term! It is:
an = [ Phin - (phi)n ]/Sqrt[5].
where Phi=(1+Sqrt[5])/2 is the so-called golden mean, and phi=(1-Sqrt[5])/2 is an associated golden number, also equal to (-1/Phi). This formula is attributed to Binet in 1843, though known by Euler before him.
The Math Behind the Fact:
The formula can be proved by induction. It can also be proved using the eigenvalues of a 2x2-matrix that encodes the recurrence. You can learn more about recurrence formulas in a fun course called discrete mathematics.
Six Degrees of Separation
The word graph has two different meanings in mathematics. One involves plotting the domain and range of a function, and another is used to model relationships between discrete objects.
In this definition, a graph is any set of vertices (dots) in which some pairs of vertices are connected by edges (lines). Often the lines are used to represent relationships between objects (represented by dots).
For example, we can construct a graph in which the vertices represent the people in this class, and we'll draw edges between any two people who mutually know one other. We can measure the "distance" between two vertices A and B by the least number of edges that one has to cross to get from A to B in the graph.
Here's a popular question: what is the minimum distance between any two people in the world, using the graph above?
It is popularly believed that the number is 6 or less for any pair of people. You may have heard the term "six degrees of separation". In fact, in the U.S. it is probably easy to get to anyone using a chain of 3 or fewer people... try using your mayor, congressman, or college professors as intermediate points!
Presentation Suggestions:
If your students like this concept, you can also mention that there is a similar concept of "Erdos number", which is the length of the chain in the graph where edges represent the relations "co-authored a paper with" and distance is measured from a famous (prolific!) number theorist named Paul Erdos. The website in the reference contains a wealth of interesting information about this relation.
The Math Behind the Fact:
Graph theory is an branch of mathematics that is very useful in computer science. You can get an introduction to graph theory in a course on discrete mathematics.
Successive Differences of Powers
List the squares:
0, 1, 4, 9, 16, 25, 36, 49, ...
Then take their successive differences:
1, 3, 5, 7, 9, 11, 13, ...
Then take their successive differences again:
2, 2, 2, 2, 2, 2, ...
So the 2nd successive differences are constant(!) and equal to 2.
OK, now list the cubes, and in a similar way, keep taking successive differences:
0, 1, 8, 27, 64, 125, 216, 343, 512, ...
1, 7, 19, 37, 61, 91, 127, 169, ...
6, 12, 18, 24, 30, 36, 42, ...
6, 6, 6, 6, 6, 6, ...
Gee, the 3rd successive differences are all constant(!) and equal to 6.
What happens when you take the 4th successive differences of 4th powers? Are they constant? What do they equal? (They're all 24.) And the 5th successive differences of 5th powers?
Aren't derivatives similar to differences? What do you think happens when you take the n-th derivative of xn?
Presentation Suggestions:
Have students do these investigations along with you. If you assign the n-th derivative of xn on a previous homework, then you can make the connection between the two right away.
The Math Behind the Fact:
This pattern may seem very surprising. It can be proved by induction. Taking differences is like a discrete version of taking the derivative, where the space between successive points is 1.
This idea has a very practical application: given a sequence generated by an unknown polynomial function, you use the calculation of successive differences to determine the order of the polynomial! Then use the first N terms of the sequence with the first N terms of the polynomial to solve for the generating function.
Birthday Problem How many people do you need in a group to ensure at least a 50 percent probability that 2 people in the group share a birthday?
Let's take a show of hands. How many people think 30 people is enough? 30? 60? 90? 180? 360?
Surprisingly, the answer is only 23 people to have at least a 50 percent chance of a match. This goes up to 70 percent for 30 people, 90 percent for 41 people, 95 percent for 47 people. With 57 people there is better than a 99 percent chance of a birthday match!
Presentation Suggestions:
If you have a large class, it is fun to try to take a poll of birthdays: have people call out their birthdays. But of course, whether or not you have a match proves nothing...
The Math Behind the Fact:
Most people find this result surprising because they are tempted to calculate the probability of a birthday match with one particular person. But the calculation should be done over all pairs of people. Here is a trick that makes the calculation easier.
To calculate the probability of a match, calculate the probability of no match and subtract from 1. But the probability of no match among n people is just
(365/365)(364/365)(363/365)(362/365)...((366-n)/365),
where the k-th term in the product arises from considering the probability that the k-th person in the group doesn't have a birthday match with the (k-1) people before her.
If you want to do this calculation quickly, you can use an approximation: note that for i much smaller than 365, the term (1-i/365) can be approximated by EXP(-i/365). Hence, for n much smaller than 365, the probability of no match is close to
EXP( - SUMi=1 to (n-1) i/365) = EXP( - n(n-1)/(2*365)).
When n=23, this evaluates to 0.499998 for the probability of no match. The probability of at least one match is thus 1 minus this quantity.
For still more fun, if you know some probability: to find the probability that in a given set of n people there are exactly M matches, you can use a Poisson approximation. The Poisson distribution is usually used to model a random variable that counts a number of "rare events", each independent and identically distributed and with average frequency lambda.
Here, the probability of a match in a given pair is 1/365. The matches can be considered to be approximately independent. The frequency lambda is the product of the number of pairs times the probability of a match in a pair: (n choose 2)/365. Then the approximate probability that there are exactly M matches is:
(lambda)M * EXP(-lambda) / M!
which gives the same formula as above when M=0 and n=-365.
Largest Known Primes Since there are infinitely many primes, what are the largest primes that we know of?
The largest known primes are ones of the form (2m - 1). The reason is that there exist efficient ways to test whether such numbers are prime. Primes of this type are called a Mersenne primes.
the largest known primes are
26,972,593 - 1
23,021,377 - 1
22,976,221 - 1
The largest is over 2 million digits long! These primes were all discovered in the last 3 years; the search for large primes has accelerated with the help of several hundred people across the internet in a project called GIMPS [the Great Internet Mersenne Prime Search]. For more on this, see the URL in the reference.
Presentation Suggestions:
Ask students to guess how large those numbers are, before you tell them.
The Math Behind the Fact:
As it turns out, knowing large primes is very important in cryptography. Being able to factor large numbers is "equivalent" to being able to crack codes, and typical codes that are nearly impossible to break are ones which depend on knowing a large number that is almost prime.
Zero to the Zero Power It is commonly taught that any number to the zero power is 1, and zero to any power is 0. But if that is the case, what is zero to the zero power?
Well, it is undefined (since xy as a function of 2 variables is not continuous at the origin).
But if it could be defined, what "should" it be? 0 or 1?
Presentation Suggestions:
Take a poll to see what people think before you show them any of the reasons below.
The Math Behind the Fact:
We'll give several arguments to show that the answer "should" be 1.
The alternating sum of binomial coefficients from the n-th row of Pascal's triangle is what you obtain by expanding (1-1)n using the binomial theorem, i.e., 0n. But the alternating sum of the entries of every row except the top row is 0, since 0k=0 for all k greater than 1. But the top row of Pascal's triangle contains a single 1, so its alternating sum is 1, which supports the notion that (1-1)0=00 if it were defined, should be 1.
The limit of xx as x tends to zero (from the right) is 1. In other words, if we want the xx function to be right continuous at 0, we should define it to be 1.
The expression mn is the product of m with itself n times. Thus m0, the "empty product", should be 1 (no matter what m is).
Another way to view the expression mn is as the number of ways to map an n-element set to an m-element set. For instance, there are 9 ways to map a 2-element set to a 3-element set. There are NO ways to map a 2-element set to the empty set (hence 02=0). However, there is exactly one way to map the empty set to itself: use the identity map! Hence 00=1.
Here's an aesthetic reason. A power series is often compactly expressed as
SUMn=0 to INFINITY an (x-c)n.
We desire this expression to evaluate to a0 when x=c, but the n=0 term in the above expression is problematic at x=c. This can be fixed by separating the a0 term (not as nice) or by defining 00=1.
Scrabble quiz Question: What positive number, when spelled out, has a Scrabble score equal to that integer?
Answer: Twelve = 12 points in Scrabble
Four Fours Problem Here's a challenge that you may wish to try: can you express all the numbers from 1 to 100 using an arithmetic combination of only four 4's?
The operations and symbols that are allowed are: the four arithmetic operations (+,x,-,/), concatenation (44 is ok and uses up two 4's), decimal points (using 4.4 is ok), powers (using 44 is ok), square roots, factorials (using 4! is ok), and overbars for indicating repeating digits (e.g., writing .4 with an overbar would be a way of expressing 4/9). Ordinary use of parentheses are allowed. No digits other than 4 are allowed.
This problem is sometimes called the four fours problem. I'm not sure where it first originated but it was popularized by Martin Gardner, among others.
Presentation Suggestions:
This puzzle makes an excellent extra credit problem. Or, you might suggest it as a joint project for a whole class to work on: have them post solutions on a bulletin board as they find them. For a computer science course it makes a nice programming exercise in the language prolog.
The Math Behind the Fact:
Actually, all the numbers less than 113 can be constructed in this fashion. While I won't spoil the fun and tell you the answers, let me just say (from experience) that the hardest numbers to express in four 4's are the numbers 69 and 73. These require especially clever combinations of the operations above.
A difficult (and as far as I know unsolved) mathematical challenge is to prove that the number 113 cannot be constructed using these operations.
It should be noted that there are many versions of this problem that have floated around, differing only in the sets of operations that are allowed. (For instance, 113 can be done if you allow arccos as a function.)
Pi Approximations Pi is the ratio of the circumference of a circle to its diameter. It is known to be irrational and its decimal expansion therefore does not terminate or repeat. The first 40 places are:
3.14159 26535 89793 23846 26433 83279 50288 41971...
Thus, it is sometimes helpful to have good fractional approximations to Pi. Most people know and use 22/7, since 7*Pi is pretty close to 22. But 22/7 is only good to 2 places. A fraction with a larger denominator offers a better chance of getting a more refined estimate. There is also 333/106, which is good to 5 places.
But an outstanding approximation to Pi is the following:
355/113
This fraction is good to 6 places! In fact, there is no "better approximation" among all fractions (P/Q) with denominators less than 30,000. [By "better approximation" we mean in the sense of how close Q*Pi is to P.]
Presentation Suggestions:
Have people verify that 355/113 is a good rational approximation. You can also point out that 355/113 is very easy to remember, since it consists of the digits 113355 in some order!
The Math Behind the Fact:
The theory of continued fractions allows one to find good rational approximations of any irrational number. This is covered in an introductory course on number theory!
Multiplication by 11 Multiplication by 11 is easy! To multiply by a 2-digit number add the two digits and place the sum in between!
25 x 11 = 275
31 x 11 = 341
57 x 11 = 627 <-- you need to carry the 1!
What about a 3-digit number? Can you figure out what's going on here?
253 x 11 = 2783
117 x 11 = 1287
532 x 11 = 5852
267 x 11 = 2937
Presentation Suggestions:
Do examples! What you notice is that multiplication by 11 can be done quickly with numbers of any length by starting with the first and last digits (they remain the same, unless there is a carry) and then inserting the sums of adjacent pairs of digits sequentially in between. For example, 253 x 11 begins with a 2 (like 253 does), then the next digit is 2+5=7, the next digit is 5+3=8, and the last digit is 3 (like 253 has). So the product is 2873. Remember to carry if necessary. So 267 x 11 starts with a 2 (like 267 does), the next digit is 2+6=8, the next digit is 6+7=13 (oops! carry the 1 back to the previous digit, leaving a 3 in this place), then the last digit is 7 (just like 267 has). Thus the product is 2937.
The Math Behind the Fact:
Long multiplication reveals why the trick works... you end up adding adjacent digits. Based on this, can you figure out a snappy rule for Multiplication by 111?
Multiplication by 111 If you liked the Fun Fact Multiplication by 11, you'll enjoy seeing how to take that idea one step farther. Here's a quick way to multiply by 111.
To multiply a two-digit number by 111, add the two digits and if the sum is a single digit, write this digit TWO TIMES in between the original digits of the number. Some examples:
23x111= 2553
41x111= 4551
The same idea works if the sum of the two digits is not a single digit, but you should write down the last digit of the sum twice, but remember to carry if needed. So
57x111= 6321
because 5+7=12, but then you have to carry the one twice.
If the number you are multiplying by 111 is a three-digit number, say ABC, then the answer will have five digits (though it may be six if there is a carry involved):
the first digit is A,
the second digit is A+B,
the third digit is A+B+C,
the fourth digit is B+C,
the fifth digit is C.
Again, you must remember to carry if any of these sums is more than one digit. Thus 123x111=13653, 241x111=26751, and for an example where carrying is needed, 352x111=39072. (Because of the carries, it may be easier to do the sums and write the answer down from right to left.)
Presentation Suggestions:
Do the Multiplication by 11 Fun Fact first.
The Math Behind the Fact:
Multiply using the traditional (long) method for multiplication, and you will find that the above shortcut works because it is doing exactly the same sums that you would have to do using the traditional method for multiplication.
Squaring Quickly You may have seen the Fun Fact on squares ending in 5; Here's a trick that can help you square ANY number quickly.
It's based on the algebra identity for the difference of squares, but with a twist! Can you figure it out?
542 = 50 * 58 + 42 = 2916.
422 = 40 * 44 + 22 = 1764.
372 = 34 * 40 + 32 = 1369.
You have to pretty proficient at multiplying one digit numbers by two digit numbers in your head to do this trick well. But if you master this, then you can build upon it in some amazing ways:
1162 = 100 * 132 + 162 = 13,200 + 256 = 13,456.
Thinking CREATIVELY about everything you learn, no matter how trivial it may seem, will allow you to find some really clever applications!
Presentation Suggestions:
If you practice this a LOT beforehand, you can start off by asking students to name any 2-digit number and you will do it in your head quickly. Then tell them the trick. But only do this with a LOT OF PRACTICE!
The Math Behind the Fact:
If you look closely, we are using the identity:
a2 = (a-b)(a+b) + b2.
The reference contains more ideas for doing fast mental calculations.
Squares Ending in 5 Give me any 2 digit number that ends in 5, and I'll square it in my head!
452 = 2025
852 = 7225, etc.
There's a quick way to do this: if the first digit is N and the second digit is 5, then the last 2 digits of the answer will be 25, and the preceding digits will be N*(N+1).
Presentation Suggestions:
After telling the trick, have students see how fast they can square such numbers in their head, but doing several examples.
The Math Behind the Fact:
You may wish to assign the proof as a fun homework exercise: multiply (10N+5)(10N+5) and interpret! The trick works for larger numbers, too, although it may be harder to do this in your head. For instance 2052 = 42025, since 20*21=420. Also, you can combine this trick with other lightning arithmetic tricks. So 1152 = 13225, because 11*12 = 132, using the Multiplication by 11 trick.
Multiplying Complementary Pairs
Quick! What's 23 x 27?
621
There's a trick to doing this quickly. Can you see a pattern in these multiplications?
42 x 48 = 2016
43 x 47 = 2021
44 x 46 = 2024
54 x 56 = 3024
64 x 66 = 4224
61 x 69 = 4209
111 x 119 = 13209
In each pair above, the numbers being multiplied are complementary: they are the same number except for the rightmost digit, and the rightmost digits add to 10.
The trick to multiplying complementary pairs is to take the rightmost digits and multiply them; the result forms the two rightmost digits of the answer. (So in the last example 1 x 9 = 09.) Then take the first number without its rightmost digit, and multiply it by the next higher whole number; the result forms the initial digits of the answer. (So in the last example: 11 x 12 = 132. Voila! The answer is 13209.)
The Math Behind the Fact:
This trick works because you are multiplying pairs of numbers of the form 10*N+A and 10*(N+1)-A, where N is a whole number and A is a digit between 1 and 9. A little algebra shows their product is:
100*N*(N+1) + A*(10-A).
The first term in the sum is a multiple of 100 and it does not interact with the last two digits of sum, which is never more than two digits long.
Difference of Squares You have all learned that
a2 - b2 = (a + b)(a - b)
But perhaps you haven't thought about how to use this to do fast mental calculations! See if you can guess how this trick can help you do the following in your head:
43 x 37
78 x 82
36 x 24
Let's do the first one. 43 x 37 = (40 + 3)(40 - 3) = 402 - 32 = 1600 - 9 = 1591.
Practice these, and you'll be able to impress your friends!
Visual Multiplication with Lines Here's a way to multiply numbers visually!
Suppose you want to multiply 22 by 13. Draw 2 lines slanted upward to the right, and then move downward to the right a short distance and draw another 2 lines upward to the right (see the magenta lines in Figure 1). Then draw 1 line slanted downward to the right, and then move upward to the right a short distance and draw another 3 lines slanted downward to the right (the cyan lines in Figure 1).
Now count up the number of intersection points in each corner of the figure. The number of intersection points at left (green-shaded region) will be the first digit of the answer. Sum the number of intersection points at the top and bottom of the square (in the blue-shaded region); this will be the middle digit of the answer. The number of intersection points at right (in the yellow-shaded region) will be the last digit of the answer.
This will work to multiply any two two-digit numbers, but if any of the green, blue, gold sums have 10 or more points in them, be sure to carry the tens digit to the left, just as you would if you were adding.
Presentation Suggestions:
First do simple examples like the one above; then try a problem that involves a carry, such as 21 x 34.
The Math Behind the Fact:
The method works because the number of lines are like placeholders (at powers of 10: 1, 10, 100, etc.), and the number of dots at each intersection is a product of the number of lines. You are then summing up all the products that are coefficients of the same power of 10. Thus the in the example
22 x 13 = ( 2*10 + 2 ) * ( 1*10 + 3 ) = 2*1*100 + 2*3*10 + 2*1*10 + 2*3 = 286.
The diagram displays this multiplication visually. In the green-shaded region there are 2*1=2 dots. In the blue-shaded region there are 2*3+2*1=8 dots. In the gold-shaded region there are 2*3=6 dots. This method does exactly what you would do if you wrote out the multiplication the long way and added the columns!
The method can be generalized to products of three-digit numbers (or more) using more sets of lines (and summing the dot groupings vertically and remembering to carry when needed). It can also be generalized to products of three-numbers using cubes of lines rather than squares! (Of course, it gets pretty unwieldy to use the method at that point.)
By the way, for the specific problem 22 x 13 there is actually another way to do it using lightning arithmetic; can you figure out how?
Making Magic Squares A magic square is an NxN matrix in which every row, column, and diagonal add up to the same number. Ever wonder how to construct a magic square?
A silly way to make one is to put the same number in every entry of the matrix. So, let's make the problem more interesting--- let's demand that we use the consecutive numbers.
I will show you a method that works when N is odd. As an example, consider a 3x3 magic square, as in Figure 1. Start with the middle entry of the top row. Place a 1 there. Now we'll move consecutively through the other squares and place the numbers 2, 3, 4, etc. It's easy: after placing a number, just remember to always move:
1. diagonally up and to the right when you can,
2. down if you cannot.
The only thing you must remember is to imagine the matrix has "wrap-around", i.e., if you move off one edge of the magic square, you re-enter on the other side.
Thus in Figure 1, from the 1 you move up/right (with wraparound) to the bottom right corner to place a 2. Then you move again (with wraparound) to the middle left to place the 3. Then you cannot move up/right from here, so move down to the bottom left, and place the 4. Continue...
It's that simple. Doing so will ensure that every square gets filled!
Presentation Suggestions:
Do 3x3 and 5x5 examples, and then let students make their own magic squares by using other sets of consecutive numbers. How does the magic number change with choice of starting number? How can you modify a magic square and still leave it magic?
The Math Behind the Fact:
See if you can figure out (prove) why this procedure works. Get intuition by looking at lots of examples!
If you are ready for more, you might enjoy this variant: take a 9x9 square. You already know how to fill this with numbers 1 through 81. But let me show you another way! View the 9x9 as a 3x3 set of 3x3 blocks! Now fill the middle block of the top row with 1 through 9 as if it were its own little 3x3 magic square... then move to the bottom right block according to the rule above and fill it with 10 through 27 like a little magic square, etc. See Figure 2. When finished you'll have a very interesting 9x9 magic square (and it won't be apparent that you used any rule)!
