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# One cool math trick (magic... !!!)

metalfreek
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 321-123=198)
4. Now reverse the digits in step 3. (ie 198 becomes 891)
5. Now, add 891 and 198.

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.
Prodigeeman
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, 201-102=99.
3. The number shouldn't end in a zero. 110-011=? 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
metalfreek
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.
cybersa

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.(15-6 = 9)
In the above magic,number divided by 9 have same symbol.
socials
OMG!! That is so amazing, how that is possible. I got king of scared when i did it twice. How did you found that?
chatrack
Hi,
Nice trick.
metalfreek
 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, 201-102=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 201-102 should be written as 099.

Now we reverse the number as 990.

Adding 990 + 99 we get 1089.
bowerryan
math teacher made us do this before. :/
infinisa
 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, 201-102=99. 3. The number shouldn't end in a zero. 110-011=? 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      b-1    10+c - c      b      a _____________________

Recalling that a = c + d, we can write:

 Code: Hundreds Tens    Units   c+d    b-1     10+c - c      b       c+d _____________________                  10-d

Here we've subtracted the units. Note that since 1 <= d <= 9, we have 1 <= 10-d <= 9, so 10-d is a bona fide digit.

To subtract the tens, we must first borrow 1 from the hundreds:

 Code: Hundreds Tens    Units   c+d-1  b+9     10+c - c      b       c+d _____________________   d-1    9       10-d

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 d-1 >= 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   d-1    9       10-d + 10-d   9       d-1 _____________________   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!
timetorock
Maths trick has always been a matter of great fun. and so is it. but would you please specify some other limitations?
rajpk
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:)
lightwate
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 0-9 for the tens place and counting down from 9-0 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
infinisa
 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.
roepjenaam
VEtcool
Arrogant
well that's really cool.
Where did you come up with that?!
cybersa
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, 201-102=99. 3. The number shouldn't end in a zero. 110-011=? 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      b-1    10+c - c      b      a _____________________

Recalling that a = c + d, we can write:

 Code: Hundreds Tens    Units   c+d    b-1     10+c - c      b       c+d _____________________                  10-d

Here we've subtracted the units. Note that since 1 <= d <= 9, we have 1 <= 10-d <= 9, so 10-d is a bona fide digit.

To subtract the tens, we must first borrow 1 from the hundreds:

 Code: Hundreds Tens    Units   c+d-1  b+9     10+c - c      b       c+d _____________________   d-1    9       10-d

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 d-1 >= 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   d-1    9       10-d + 10-d   9       d-1 _____________________   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.
tyler87898
This is quite interesting. Hope I can remember it.
WolfAngelxx
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.
kelseymh
 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.
ajaybe
OMG!! That is so amazing, how that is possible.
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