
All right guys here is a cool maths magic that you can try out with your friends. Remember that this trick should be done only once with one person.
1: Take a three digit numbers. (eg: 123)
2: Now reverse the digits. (reverse of 123 is 321)
3: Now, subtract smaller number from bigger number. (in our case 321123=198)
4. Now reverse the digits in step 3. (ie 198 becomes 891)
5. Now, add 891 and 198.
6. The answer is 1089
The answer is always 1089 no matter what three digits numbers you choose.
But before you start to show this tricks make sure that you know about the limitations. This trick doesn't work if you choose 101 as the starting number.
There are others limitation to this trick. Make sure you know about that before showing off the magic.
This was a little amusing, but still fun to figure out!
So I found these limitations in going through the sequence a few times:
1. No symmetrical numbers, such 606, or 747. They can't be reversed.
2.There can't be any numbers in which when reversed, would subtract to get a symmetrical number, or a number lower than 3 digits. For example, 201102=99.
3. The number shouldn't end in a zero. 110011=? If we say 11, the answer is still 99.
4. The number should not end with a number that is within one integer of the beginning number. For example, 695. 5 is within 1 integer of 6, and when solved through the answer is again 99. This is because you are subtracting a number that was forced to be within one hundred after you reversed the first number.
In finding these rules, I also noticed that each sequence ends up to be either 99 or 198 after step 3. I wonder if there is one that will subtract to 297? There has to be extraneous ones though.. Large differences between the 1st and 3rd digit would produce those results.
Also, who can find the two rules that go together? One was built upon the other, and there are only 4 rules, so it shouldn't be too hard
Just choosing a three digit number will be good for this trick. The limitation is a risk in part of the performer. But violating these rules will be less I guess.
Thanks for your trick.
Another Trick:
Here is the magic:http://www.exstatica.net/flash/psychic.swf
First try above.
Solution
1.Take any two digit number from 10 to 99(Example:15)
2.Then separate into two digit and add it (1+5=6)
3.Then subtract the chosen number with answer.(156 = 9)
Your final answer is always divided by 9.
In the above magic,number divided by 9 have same symbol.
OMG!! That is so amazing, how that is possible. I got king of scared when i did it twice. How did you found that?
Prodigeeman wrote:  This was a little amusing, but still fun to figure out!
So I found these limitations in going through the sequence a few times:
2.There can't be any numbers in which when reversed, would subtract to get a symmetrical number, or a number lower than 3 digits. For example, 201102=99.

You are right but when 2 digits pops up what you can do is add one zero in front of the number. In your case 201102 should be written as 099.
Now we reverse the number as 990.
Adding 990 + 99 we get 1089.
math teacher made us do this before. :/
Prodigeeman wrote:  This was a little amusing, but still fun to figure out!
So I found these limitations in going through the sequence a few times:
1. No symmetrical numbers, such 606, or 747. They can't be reversed.
2.There can't be any numbers in which when reversed, would subtract to get a symmetrical number, or a number lower than 3 digits. For example, 201102=99.
3. The number shouldn't end in a zero. 110011=? If we say 11, the answer is still 99.
4. The number should not end with a number that is within one integer of the beginning number. For example, 695. 5 is within 1 integer of 6, and when solved through the answer is again 99. This is because you are subtracting a number that was forced to be within one hundred after you reversed the first number.
In finding these rules, I also noticed that each sequence ends up to be either 99 or 198 after step 3. I wonder if there is one that will subtract to 297? There has to be extraneous ones though.. Large differences between the 1st and 3rd digit would produce those results.
Also, who can find the two rules that go together? One was built upon the other, and there are only 4 rules, so it shouldn't be too hard 
Hi Prodigeeman
Here is the explanation:
Take a general 3 digit number, say a$b$c (here the $ means concatenation, e.g. 1$2$3 = 123).
We then reverse it, to get c$b$a.
Now we subtract the smaller from the larger.
Let's suppose, for now, that a>=c (we'll look at the other case later).
Then we get a$b$c  c$b$a.
If a=c, we get 0, so rule this out i.e. (Rule 1) exclude numbers like a$b$a (symmetrical numbers).
So (still assuming a >= c) we have a > c, and can write a = c + d where 1 <= d <= 9.
Let's now do the subtraction:
Code: 
Hundreds Tens Units
a b c
 c b a
_____________________

Units: subtract a from c. But a > c, so we must borrow 1 from the tens, and get:
Code: 
Hundreds Tens Units
a b1 10+c
 c b a
_____________________

Recalling that a = c + d, we can write:
Code: 
Hundreds Tens Units
c+d b1 10+c
 c b c+d
_____________________
10d

Here we've subtracted the units. Note that since 1 <= d <= 9, we have 1 <= 10d <= 9, so 10d is a bona fide digit.
To subtract the tens, we must first borrow 1 from the hundreds:
Code: 
Hundreds Tens Units
c+d1 b+9 10+c
 c b c+d
_____________________
d1 9 10d

Now lets add this result to its reverse.
To get what we want, we need this to be a genuine 3 digit number: (Rule 2) if a > c, then d1 >= 1, i.e. d >= 2, i.e. a  c >= 2.
i.e. if a > c then a >= c + 2, i.e. a must be at least 2 greater than c (this rules out 201 & 110).
Code: 
Hundreds Tens Units
d1 9 10d
+ 10d 9 d1
_____________________
10 8 9

Note that the tens add up to 18, carrying 1 to the hundreds.
We get the expected answer: 1089
Remember that we assumed a >= c.
So let's now look at the case a < c.
Then once again we need a genuine 3 digit number, so (Rule 3) a <> 0
the above rules apply to the reversed number, c$b$a
Rule 1 (symmetrical numbers) can't happen.
Rule 2 tells us (Rule 4) that if c > a then c >= a + 2, i.e. c must be at least 2 greater than a (this rules out 102).
So we have 4 rules (swapping the numbering of 2 & 3):
1) a <> c
2) a <> 0
3) if a > c then a >= c+2, i.e. a<>c+1
4) if c > a then c >= a+2, i.e. c<>a+1
So there you have it, the whole book of rules, no more unruly numbers lurking out there!
Maths trick has always been a matter of great fun. and so is it. but would you please specify some other limitations?
good tricks
i haven't any trick to share right now:(
but learn lot of tricks
hopefully i will share it when i have one
thanks:)
good trick.
I think this has something to do with the property of the multiples of the number nine. For example, I noticed that making a table for multiples of 9 is simply counting up from 09 for the tens place and counting down from 90 for the one's place:
0 9
1 8
2 7
3 6
4 5
5 4
6 3
7 2
8 1
9 0
The first is just the reverse of the last, and so on...
Though I can't explain the rest lol. But I think the rest have something to do with the number 11 and 111
timetorock wrote:  Maths trick has always been a matter of great fun. and so is it. but would you please specify some other limitations? 
Hi timetorock
If you believe my working (and I do, more or less...) then there are no other limitations.
VEtcool
well that's really cool.
Where did you come up with that?!
infinisa wrote:  Prodigeeman wrote:  This was a little amusing, but still fun to figure out!
So I found these limitations in going through the sequence a few times:
1. No symmetrical numbers, such 606, or 747. They can't be reversed.
2.There can't be any numbers in which when reversed, would subtract to get a symmetrical number, or a number lower than 3 digits. For example, 201102=99.
3. The number shouldn't end in a zero. 110011=? If we say 11, the answer is still 99.
4. The number should not end with a number that is within one integer of the beginning number. For example, 695. 5 is within 1 integer of 6, and when solved through the answer is again 99. This is because you are subtracting a number that was forced to be within one hundred after you reversed the first number.
In finding these rules, I also noticed that each sequence ends up to be either 99 or 198 after step 3. I wonder if there is one that will subtract to 297? There has to be extraneous ones though.. Large differences between the 1st and 3rd digit would produce those results.
Also, who can find the two rules that go together? One was built upon the other, and there are only 4 rules, so it shouldn't be too hard 
Hi Prodigeeman
Here is the explanation:
Take a general 3 digit number, say a$b$c (here the $ means concatenation, e.g. 1$2$3 = 123).
We then reverse it, to get c$b$a.
Now we subtract the smaller from the larger.
Let's suppose, for now, that a>=c (we'll look at the other case later).
Then we get a$b$c  c$b$a.
If a=c, we get 0, so rule this out i.e. (Rule 1) exclude numbers like a$b$a (symmetrical numbers).
So (still assuming a >= c) we have a > c, and can write a = c + d where 1 <= d <= 9.
Let's now do the subtraction:
Code: 
Hundreds Tens Units
a b c
 c b a
_____________________

Units: subtract a from c. But a > c, so we must borrow 1 from the tens, and get:
Code: 
Hundreds Tens Units
a b1 10+c
 c b a
_____________________

Recalling that a = c + d, we can write:
Code: 
Hundreds Tens Units
c+d b1 10+c
 c b c+d
_____________________
10d

Here we've subtracted the units. Note that since 1 <= d <= 9, we have 1 <= 10d <= 9, so 10d is a bona fide digit.
To subtract the tens, we must first borrow 1 from the hundreds:
Code: 
Hundreds Tens Units
c+d1 b+9 10+c
 c b c+d
_____________________
d1 9 10d

Now lets add this result to its reverse.
To get what we want, we need this to be a genuine 3 digit number: (Rule 2) if a > c, then d1 >= 1, i.e. d >= 2, i.e. a  c >= 2.
i.e. if a > c then a >= c + 2, i.e. a must be at least 2 greater than c (this rules out 201 & 110).
Code: 
Hundreds Tens Units
d1 9 10d
+ 10d 9 d1
_____________________
10 8 9

Note that the tens add up to 18, carrying 1 to the hundreds.
We get the expected answer: 1089
Remember that we assumed a >= c.
So let's now look at the case a < c.
Then once again we need a genuine 3 digit number, so (Rule 3) a <> 0
the above rules apply to the reversed number, c$b$a
Rule 1 (symmetrical numbers) can't happen.
Rule 2 tells us (Rule 4) that if c > a then c >= a + 2, i.e. c must be at least 2 greater than a (this rules out 102).
So we have 4 rules (swapping the numbering of 2 & 3):
1) a <> c
2) a <> 0
3) if a > c then a >= c+2, i.e. a<>c+1
4) if c > a then c >= a+2, i.e. c<>a+1
So there you have it, the whole book of rules, no more unruly numbers lurking out there! 
Nice explanation.
Thanks bro.
This is quite interesting. Hope I can remember it.
Now I tried to use 808, but it didn't work out the way it should have and instead, it gave me a wrong answer... Otherwise, my friends might find this more amusing.
WolfAngelxx wrote:  Now I tried to use 808, but it didn't work out the way it should have and instead, it gave me a wrong answer... Otherwise, my friends might find this more amusing. 
I guess you didn't read any of the other posts in this forum topic, including the second one which made exactly that observation.
OMG!! That is so amazing, how that is possible.
