# Math tricks

### #1

Posted 04 February 2006 - 10:25 AM

1) Take any 3 digit number, But the first and third number Cannot be the same.

853

2) reverse the order and subtract

853

-358

=495

3) reverse the answer and then add it

495

+ 594

=1089

The Answer always equals 1089!

try again: 921

-129

= 792

792+297= 1089!

Amazing huh?

Got an explanation for this?

Have any other Math Tricks of your own?

### #2

Posted 04 February 2006 - 10:28 AM

Try it with 102Heres a Math trick someone showed me the other day.

1) Take any 3 digit number, But the first and third number Cannot be the same.

### #3

Posted 04 February 2006 - 10:35 AM

It didn't take you long to debunk my trickery!

### #4

Posted 04 February 2006 - 11:01 AM

Ask someone to think of a 3 digit number where none of the digits are the same.

Have them reverse the number and subtract the smaller from the larger.

Have them tell you the last digit of the answer and you tell them the whole answer.

This sequence will always yield an answer where the middle digit is 9 and the sum of the first and last digits is 9. The only exception will be when they select an initial three digit number where the first and last digits only differ by 1. In this case the final answer will be 99.

### #5

Posted 04 February 2006 - 11:51 AM

A little bit of simple algebra can explain this, and many other similar, tricks. Here goes:Take any 3 digit number…

Got an explanation for this?

*Take any 3 digit number, But the first and third number Cannot be the same.*

Let’s represent our 3 digit number as 100A +10B +C*reverse the order and subtract*

Gives 100A +10B +C – (100C +10B –A) =

99(A -C)

Now a little intuition is needed. Lets substitute “D +1” for “A –C”, rewrite the number as

99(D +1)

and rearrange this into

99D + 99 =

100D -D +90 +9 =

100D +10(9) +(9 -D)*reverse the answer and then add it*

Gives 100D +90 +(9 -D) +(100(9 -D) +90 +D),

which rearranges into

100(D +(9 -D)) +180 +(9 - D +D) =

100(9) +180 + 9 =

1089

- Jay-qu likes this

### #6

Posted 06 February 2006 - 07:56 AM

Maybe it can be said more simply. The first step clearly doesn't depend on the central digit and it will be 99 times the difference between the left and the right digit. This is negative in C1ay's case.

The three digit reverse of 099 is 990 and their sum is 1089 (-1089 in C1ay's case). Each time you add 99 to the first, the second is decreased by 99. Why? The reason for this is the trickiest part, you are adding 990 with a backward carry rule, which in the case of the digit 9 amounts to the borrow rule for subtracting 9 from the digit to the right, because both these things decrease the one digit and increase the other.

99 + 99 reversed is 891 = 990 - 99 etc.

### #7

Posted 06 February 2006 - 11:20 AM

Math is logical, even when parlor tricks and games are involved.

Anyone have any quirky and fun math tricks?

### #8 Guest_chendoh_*

Posted 28 February 2006 - 02:12 PM

I've been trying to get it to work for succeeding years by changing numbers 2,3,4,

and 5, mostly #5 with no luck. maybe with a little help from our resident math wizards.

Though, it will tell a persons age in 2002.

This is pretty neat how it works out.

DON'T CHEAT BY SCROLLING DOWN FIRST

It takes less than a minute.......

Work this out as you read.

Be sure you don't read the bottom until you've worked it out!

This is not one of those waste of time things, it's fun.

1. First of all, pick the number of times a week that you would like

to have dinner out. (try for more than once but less than 10)

2. Multiply this number by 2 (Just to be bold)

3. Add 5. (for Sunday)

4. Multiply it by 50 - I'll wait while you get the

calculator................

5. If you have already had your birthday this year add 1752.... If

you haven't, add 1751..........

6. Now subtract the four digit year that you were born.

You should have a three digit number .

The first digit of this was your original number

(I.e., how many times you want to have eat out each week.)

The next two numbers are .

YOUR AGE! (Oh YES, it IS!!!!!)

THIS IS THE ONLY YEAR (2002) IT WILL EVER WORK, SO SPREAD IT AROUND WHILE IT LASTS. IMPRESSIVE, ISN'T IT?

### #9

Posted 28 February 2006 - 08:41 PM

This is an easy one to figure out with some simple algebra. Just write the 6 instructions as an expression… Here's one that I enjoyed, back in 2002.

I've been trying to get it to work for succeeding years by changing numbers 2,3,4,

and 5, mostly #5 with no luck. maybe with a little help from our resident math wizards.…

50(2n+5) +1752 –y

, where n is the number from step 1, and y is your birth year

, simplify it to

100n +2002 –y

. 2002 –y is just the usual formula for your age in 2002. As long as you’re under 100 years old, and n is from 1 to 9, you’ll get a 3-digit number, where the last 2 are your age.

You can make the problem work for any year by changing the step 5 numbers to the year -250 and year -249, or you can change the numbers in all the steps, as long as the step 2 and step 4 number multiplied are 100, and the step 5 number is the current year –(step 3 number)*(step 4 number). For example

20(5n +17) +1666 –y

translates into

1) pick a number from 1 to 9

2) multiply it by 5

3) add 17 to it

4) multiply by 20

5) add 1666 if you’ve already had your birthday, 1665 if you haven’t

6) subtract the four digit year that you were born.

With a little algebra, you can create countless tricks like this

### #10

Posted 17 January 2010 - 08:19 AM

try it with 102

Edit: Oops! Did not se Qfw's post above.

### #11

Posted 17 May 2010 - 09:22 AM