saltfish

I am aware of a few interesting mathematical tricks...

The Sum of Nine: Any number multiplied by nine will have its digits up to nine.

Example:
8*9 = 72 | 7+2=9
16*9 = 144 | 1+4+4=9
256*9 = 2304 | 2+3+0+4=9
16384*9 = 147456 | 1+4+7+4+5+6= 27 | 2+7=9

This works with ANY whole number multiplied by 9.

I am also aware of a equation that will enable you to determine what day of the week a particular date will fall on...

January 20th 1886 = XXXday..

I don' know the specifics and hope that maybe one of our resident mathemeticans will have a clue.

Anyways, if use any little math tricks and are willing to share, I know that I'd like to hear them.

-SF

KnifeMissile

I'm more interested in why these mathematical tricks work than their mere existence.
For instance, I've been thinking about this whole "sum of 9's" thing and I think I finally see how it works.

9 is the last digit in the decimal system, so when you add another 9 to it, you necessarily increment the next digit and decrement the current digit. So:

2 * 9 = 18
3 * 9 = 27
4 * 9 = 36

...you get the idea. Because of this digit conservation (one digit place decreases while the other increases by exactly the same amount), they must necessarily sum to 9! (exclamation, not factorial!)

Now, you're probably thinking "but what about when the digits sum to a number greater than nine!" Well, you'll note that they sum to a multiple of nine! This is necessarily true, because of how our digital representation of numbers (yes, numbers do exist beyong the written notation we use to represent them), like so:

let b = 10 (the base)

16384 = b^4 + 6*b^3 + 3*b^2 + 8*b + 4

...now, if you were to multiply 16384 by 9, you will be multiplying each term by 9 (the distributive property of multiplication). But then each term will be made of digits that will sum to 9 because they are only single digit numbers, and we know that those digits will sum to 9, so all these digits put together must sum to some multiple of 9! (exclamation...) We all know what property multiples of 9 have, right? Their digits sum to nine!

Actually, this last part is a bit of circular logic since we're trying to prove that summing the digits of multiples of nine sum to nine but, believe it or not, it's the start of a rigorous proof, because we know that the sum of a number's digits must be less than the number. So, you can use a form of mathematical induction called (if memory serves me) strong induction. Anyway, I hope you all get the point!

Furthermore, because we understand how this works, we can safely say that this property really isn't all that weird. For instance, if you work with hexadecimal numbers, you'll note that the same thing happens with the number F (fifteen). That is, multiply any number with fifteen and then sum the hexadecimal digits of that product. You will get a number that is a multiple of fifteen! Coincidence? No, because 15 is 1 less than the hexadecimal base, 16. Try 7 base 8 (octal). As long as the digit is one less than the base, this trick will work in that base...

Pragma

My mathematical trick (caveat: I did not invent this, I discovered it while reading a book about Richard Feynman, who learned the trick from some fellow scientist/mathematician, whose name escapes me).

To calculate the square of numbers (especially those near 50):
Given a number X, that you want to find the square of:
50^2 = 2500.
Add to this number (X-50)*100 (ie: for X = 48, add -200, for X = 53, add 300).
This will give you the "almost exact" answer.
You'll have, for example, 2300 is the square of 48. However, if you need to be exact, just add (X-50)^2.

Sorry if I didn't explain it too well, but it's an interesting trick that, if you do it enough, you can quickly churn out squares on the spot.

__________________
Eat antimatter, Posleen-boy!

Giltwist

Pick any repeating decimal, for this example, lets use .2727272727...

First figure out the number of digits that repeat, in this case 27 is repeated over and over again, so that would be 2. Now, raise ten to teh number you just found, so you now have 100. Here is something neat, 100 times that original number gives you 27.2727272727... But look at the decimal portion of that number, its the number we started out with, so we subtract the original number, so we have 100 times the number minus the number, thats 99 times the number. So, dividing both sides by 99, our number is 27/99. Magical. With a little finagling, this works for any repeating decimal.

saltfish

You guys have come out with some great explanations and tricks, Giltwist, I'm going to experiment with yours it sound sreally helpful and time saving.

-SF

Giltwist

Heh, I am a math major, what do you expect? If you think that is cool, I can show you guys how to do fraction approximation of any decimal instead of exact fractions for repeatings.

Dwarf020

my dad's math trick was very basic. if you add two numbers, and notice if they're odd or even, you can predict the outcome's oddness or evenness. This is amazing to a little kid. so, to prove it:

if you add two even numbers, you get an even answer.
ex. 2+2=4, 2+4=6, 60+18=78, etc.

if you add two odd numbers, you also get an even answer.
ex. 1+3=4, 17+3=20, 21+201=222, etc.

on the other hand, if you add and odd and even number, you get an odd answer.
ex. 1+2=3, 21+6=27, 15+3=18, etc.

if you tell this one to a little kid, of the right age, they will then spend the next however long trying to prove it wrong.

wlcm

i believe pragma's trick is just a linear approximation of x^2 around x = 50.

like take the derivate at the point and use it in the point-slope forumla.

Sleepyjack

After i read 1984, i did this little equation based on another simple equation my dad showed me. It has the same trick, but if you haven't seen it before, it mught catch you by surprise

2+2=5

1. let -> x = 2 + 2
2. x(x-1) = (2+2)(x-1) (multiply by x-1)
3. x^2 -x = 2x + 2x - 2 - 2
4. x^2 - 4x = x - 4 (subtract 3x)
5. (x-5)(x^2-4x) = (x-5)(x-4) (multiply by x-5)
6. x^3 - 9x^2 + 20x = x^2 - 9x + 20
7. x(x-4)(x-5) = (x-4)(x-5) (factor)
8. x^2 - 5x = x - 5 (divide by x-4)
9. x^2 - x = 5x - 5 (add 4x)
9. x(x-1) = 5(x-1) (factor)
10. x = 5 (divide by x-1)

so 2+2=5 and all else follows.....

Giltwist

Can't divide by (x-4) or (x-1) because in both cases it is possible that term is zero. In fact, seeing how you defined x, (x-4) IS zero.

Pragma

I think he realized that, Giltwist, and that was the point of the joke - people who don't understand the math go "... whoa", and those of us who understand math shake our heads and laugh.

A professor of mine did a formal proof in class of 2=1, got most of the class believing that the laws of math were that easily broken. Then he told them the truth.

__________________
Eat antimatter, Posleen-boy!

KnifeMissile

Actually, there's nothing strictly wrong with "dividing by zero". In other words, it's quite possibly for 0 to have a multiplicative inverse and, yes, that does imply that 2 = 1, if you define 2 = 1 + 1. However, if you were to assume that 2 != 1, then there can be no multiplicative inverse and, hence, no "division by zero".

How's that for exercising logic?

saltfish

You guys have a lot of great tricks! Thank you and keep it up!

-SF

Artsemis

That's one of the methods used to proove that .9.... (repeating) = 9/9 = 1

__________________

- Artsemis
~~~~~~~~~~~~~~~~~~~~
There are two keys to being the best:
1.) Never tell everything you know

Giltwist

KM: But we do define 2!=1. I believe next semester I do the proof that 1+1=2. I seem to recall my older friends mentioning something called the successor function

KnifeMissile

Is this a rhetorical question? I can't hlep but notice the lack of a question mark...

Anyway, I was referring to the fact that the only field where the additive identity (zero) has a multiplicative inverse (dividing by zero) is the trivial field, where 1 + 1 = 2 = 1.

Giltwist

Don't forget the null field.

QuasiMojo

just stop

: )

__________________
AzAbOv ZoBeLoE

nash

Here's a "trick":

When multiplying two two-digit numbers "ab" and "cd" (where "a" and "c" are the tens digits and "b" and "d" are the ones digits of the respective two-digit numbers):

Step 1: Add the products of a*d and b*c and multiply by 10 (i.e., add a zero to the end of that)
Step 2: Multiply b*d and add that to the result in Step 1.
Step 3: Multiply a*c*100 and add that to the result in Step 2.

Example: 8 is "a", 3 is "b", 4 is "c", 2 is "d"
83
x42

Step 1: (8*2+3*4)*10 = 280
Step 2: 3*2 + 280 = 286
Step 3: 8*4*100 = 3200; 3200+286=3486 <-- Final Answer

Basically multiply the diagonals, add them together, multiply the columns add to previous results. A lot of the time, I find it easier to do this in my head since there are less numbers to keep track of. My friend's dad figured this out (by himself!) and taught us when we were in elementary school. There's probably a similar method for three digit numbers. If anyone figures it out, please enlighten us!

Corneo

Back to the nine thing. Digits of mulitples of 9 alternate.
9
18
27
36
45
54
65
72
81

See how the 1 and 8 switch places, the 2 and 7 switched...

Sapper

Search for: Doomsday calculation

It was only good up until 1999

__________________
rm -f /bin/laden

metalgeek

kinda mathy.
to figure out using compound interest how long it will take your money to double, divide the percent interest your getting by 72.
so
8% interest will take 9 years to double.
ahh the joys of compound interest.

the other thing I do, I don't know how much of a trick it is, and how useful it would be to most people is in multiplying large numbers.

45X182
I'd do 40X100 to give 4000
then 5 X10 to give 4500
40X80 to give 7700
5X80 for 8100
40x2 for 8180
5X2 fo 8190
works really well for estimating. you also have to be able to keep multiple numbers in your head. often I don't bothing thing 40X100=4000 I'd tjust think 4and then keep adding numbers to it based on if theyd go before or after. but thats alot harder to explain

__________________
The truth is, wherever you choose to be, it's the wrong place.
Chuck Palahniuk , Diary

DonnChadh

nash, that multiplication trick is great

I nearly always have a calcuator to hand, so i've lost the knack how to do stuff like that in my head, which is a pain when i dont have a calc nearby

__________________
I try to take life one day at a time, but sometimes several days attack me at once.

Outlander36

Phone number trick, VERY cool:

Ignore your area code. Use only 7 digit phone no.:

1) key-in the first 3 digits of your phone number into the calculator
2) multiply by 80
3) then plus 1
4) multiply by 250
5) plus last four digit of phone number
6) plus last four digit of phone number again
7) minus 250
8) divide by 2 at last

Is the answer your phone number???

HamiC

One more version of the "nine thing"....

To find the product of 9 times any single digit X greater than 1, simply subtract 1 from X and place the result in the tens place while placing the difference between the tens place value and 9 in the ones place.

For example, with X=8....

9 * 8 = _ _

(8-1=7)

9 * 8 = 7 _

(9-7=2)

9 * 8 = 72

Not sure why I always found this one interesting/useful. Perhaps it is because I have a hard time memorizing certain things and used this to get through my multiplication tables back in 3rd grade. Yes. It would seem that the formula would be harder to remember than the answers, but my memory has never quite worked that way.

sky_driver

Cool telephone number trick Outlander.

shakran

also, back to the 9 trick:

For numbers 0-10, whatever you multiply by 9, subtract 1 from it and you have the first number. 11-20, subtract 2. 21-30, subtract 3, and so on. You now have all but the last number of the answer. The final number is whatever makes the string of numbers add up to 9. This works up to 150.

ie.

9x6:

6-1 = 5

first number = 5
5+4=9

answer = 54


longer example: 9x150

150-15 = 135

1+3+5=9 so the final digit is 0.

9x150=1350

btw, the "digits add up to 9" trick sort of breaks down over 150 for awhile, and the digits add up to 9x2, or 18:

151x9 = 1359

1+3+5+9=18

185x9=1665
1+6+6+5 = 18

etc etc

shAzb0t

Very cool, Outlander. Never seen that one before.

KnifeMissile

This is no more amazing than simply taking the first three digits of your number and multiplying by 10000 and then adding the last four digits because that's what these operations simplify to if you were to do them. Of course, doing this is no more amazing than simply keying in your phone number into your calculator.
Did you make this up or did you hear of this somewhere?

TheBrit

Multiply any number by 11!
For example, 11
1) copy down the first number
1
2) add the 2nd number to the number on its right, and put that number to the left of the first number in the answer.
21
2) copy down the end number, put that to the left
121
Taadaa.

It works for longer numbers as well, just go in steps, left to right, and add the numbers to the number on its right
14723
161953
If you come across a number >9, for example 4 + 7 = 11 in the last example, write a 1 and add a 1 to the next number you write down.

I have a whole book of these, "The Trachtenberg Speed System of Basic Mathematics" The guy is a genius.
http://www.amazon.com/exec/obidos/tg...glance&s=books
Get that book. You can work out square roots of numbers in your head in less than a minute, etc. Not even hard work, I read and understood it when I was 11.