The Magic Number➖➕by Howard Thurston #mentalism

[MarcosNaceli]

Hey people, note this:
* Any 3 *all different* numbers. Including 0.
* After the first substraction: ignore the negative (-) and consider any two digit number as a three digit number filling with a zero.
Example 1: 506-605=-99 change -99 to 099. Its reverse is 990. Finally: 099+990=1089.
Example 2: 012-210=-198. change it to 198. Finally: 198+891=1089.
Have fun! :)

[timlonsdale]

Represent the number as: 1*X + 10*Y + 100*Z.
The inverse would be: 1*Z + 10*Y + 100*X.
Subtract the inverse and get: 99*Z - 99*X.
Also can be written as: 99*(Z-X).
Lets replace (Z-X) with N: 99*N.
Lets rewrite as: 100*N - N.
Lets rewrite as 100*(N-1) + 100 - N.
Lets rewrite as 100*(N-1) + 90 + (10-N).
Thus the inverse is: 100*(10-N) + 90 + (N-1).
Add these together and we get: 99*N + 1000 - 100*N + 90 + N - 1.
Squish together and get: 1000 + 90 - 1 = 1089

QED

[Todd-rq]

I've watched all your videos .I love magic. Since I was 15.57 now...your very beautiful. I learned a lot. Don't stop practicing magic..

[doomdoomtv]

Showed this to my step dad. He very audibly said what the fuck and that was very surreal to hear from the man who has never said anything worse than "bugger". Thank you, incredible as always

[nicemanx]

Pro tip - don't ask same person twice.

[muhammad]

If abc is the 3 different digit number we can write it algebraically as 100a+10b+c
Then the reverse digit number will be 100c+10b+a
Now subtract the reverse digit number from original one
(100a+10b+c)-(100c+10b+a)
The new number will be
(100-1)a+(10-10)b+(1-100)c
Or
99a-99c
Or *99(a-c)* (1)
Since a and c are single digit numbers i.e
a, b belongs [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
So (a-c) will also be a single digit number.
[But a≠0 because abc is a 3 digit number]
Henc (a-c) belongs to [1, 2, 3, 4, 5, 6, 7, 8, 9]
_We excluded 0 because
a-c=0 if and only if a=c
And we know that a, b and c are three different digits. a≠c

Now lets substitute the other possible values of (a-c) one by one
For *a-c=1*
(1)=> 99×1=99
Now add reverse
99+99=198
Add reverse digit number again
198+891=1089
For *a-c=2*
(1)=> 99×2=198
Now add reversed digits
198+891=1089
For *a-c=3*
(1)=> 99×3=297
Now add reversed digits
297+792=1089
For *a-c=4*
(1)=> 99×4=396
Now add reversed digits
396+693=1089
For *a-c=5*
(1)=> 99×5=495
Now add reversed digits
495+594=1089
For *a-c=6*
(1)=> 99×6=594
Now add reversed digits
594+495=1089
For *a-c=7*
(1)=> 99×7=693
Now add reversed digits
693+396=1089
For *a-c=8*
(1)=> 99×8=792
Now add reversed digits
792+297=1089
For *a-c=9*
(1)=> 99×9=891
Now add reversed digits
891+198=1089
Similarly for negative values of (a-c)
Ignore the negative we will conclude with the same answers.
I have tried to verify it by putting all possible values. I will be more than grateful if you share your kind opinion or correction.

[ShouryaYeralkar]

Legends who got 2 same numbers like 99 and couldn't reverse it

[fschorn]

It's really a stretch to make it work when the difference is 99. For example 231-132=99. You'd need to see the 99 as 099, then reverse to 990. Adding back 990 to 99 does get you 1089.

[indianvfxschool]

Here's why it happens:

1. **Three-Digit Number**: Consider any three-digit number where the first and last digits differ by at least two. Let's call it ABC, where A, B, and C are its digits.

2. **Reversing and Subtracting**: When you reverse this number, you get CBA. Subtracting CBA from ABC (assuming A is greater than C) gives a new number. This subtraction aligns with the form (100A + 10B + C) - (100C + 10B + A) = 99A - 99C.

3. **Result of Subtraction**: Simplifying 99A - 99C, you get 99(A - C). Since A and C are at least two digits apart, the minimum value for (A - C) is 2 (e.g., if A is 3 and C is 1), and the maximum is 9 (e.g., if A is 9 and C is 0). This results in a number that is two digits long, and its first digit is always 9.

4. **Intermediate Number Characteristics**: The intermediate number (after subtraction but before adding the reverse) is always of the form 9X9, where X is a digit. This is because the hundreds and units place will always be 9 due to the subtraction process described above.

5. **Reversing the Intermediate Number**: When you reverse this number, it becomes 9X9 again, as reversing doesn't change the position of the digits due to its symmetry.

6. **Final Sum**: Adding the intermediate number and its reverse, both being 9X9, will always lead to 1089. This is because 900 + 90 + 9 equals 999, and adding it to another 999 (since the reverse is the same) results in 1089.

In summary, the specific properties of base-10 subtraction and the constraints placed on the original number (three digits, first and last digits differ by at least two) ensure that this trick works, leading to the consistent result of 1089.

[dan_marko]

Mathematical explanation:

Let's say that the hundreds digit is greater than the units digit (if its the other way you will need to negate the result which is basically like negating the first operation and -(x-y)=y-x so we can continue as if it was this way to begin with) and let's call the hundreds digit "a" and the units digit "b" so a>b.

I will also represent the numbers with their digits divided by commas so I would be able to represent digits with algebraic expressions (132=1, 3, 2)

Step #1: subtract the reverse:

Let's separate the subtraction into steps:
A. Subtract the middle digit: the middle digit will be subtracted by itself so it does not matter. We are left with a, 0, b - b, 0, a
B: Subtract the hundreds digit: as a>b we know that the result will only affect the hundreds digit. Now we are left with (a-b), 0, b - a
C: Subtract the units digit: as b<a we would need to carry over one from the tens digit, which is 0, so we need to also carry over one from the hundreds.
We are finally left with: (a-b-1), 9, (10+b-a)

Step #2: add the new reverse:

we can write the new number as one algebraic expression like so: 100*(a-b-1)+10*9+(10+b-a)
and the reverse like so:
100*(10+b-a)+10*9+(a-b-1)
Let's add them:
100*(a-b-1)+10*9+(10+b-a) + 100*(10+b-a)+10*9+(a-b-1) =

=100*9+10*18+9=900+180+9=1089

So for every 3 digit number with 3 different digits, this algorithm would yield 1089 (actually the middle digit can be equal to one other digit but if the hundreds and the units digits are the same you'd get 0)

[benjaminfischer]

I love the way you say "let's learn". I only wish my teachers in school would have encouraged me like you have.

[Чернолесье]

Спасибо, это было очень достойно для фокусов с числами 👍

[RJS-xu]

When I was 8 I discovered that you can take any number larger than 9 and up to infinity and subtract its opposite and that number is ALWAYS perfectly divisible by 9, or 3 because 3 squared is obviously 9

[bogdanmikita]

453 - 354 = 99 to in some cases where there is a difference of 1 between first and last digit you would have to add 0 to the first result to get 1089 but it will still work just add extra step by saying: " Add 0 in the end if you got a two digit result".

[TrainsandRockets]

This was discovered by Ramanujan while sitting in a taxi...
The Math Genius of the millennia...

[bhagawanpawar]

Glad to see someone else know this,
But some addition to this is
Surprisingly it is 33*33

Also this no (33*33) = 1098 is exactly used to represent 1 Guntha area in terms of square feet.

Also if you want to make it in bigger version
That is for some more digits number ( instead of 3 suppose we take n digit number)

Then the term of reverse the number changes to interchange to first and last number in subtraction and addition
You will get

10[( n-3)times 9] 89

I.e. for 5 digit number answer is 109989

E.g.
65731
- 15736
=49995
+59994
=109989

❤ Thank you

[dnxsol]

Very nice, this will score many free drinks at the pub for me here in Ireland. Thanks for always offering easy doable tips and tricks.
Love the channel

[Groundhog]

I paint cars for one job, but I work in a bar for my second job and this simple thing has blown a few minds. It’s fun to see the reaction on new customers. It’s a great way to generate fun conversation and I don’t know how many will start watching your channel, but the comments I get have been fantastic. I’ve been using a few card tricks and it’s brought in new people as well. Thank you so much for teaching me. Magic brings the inner child out in everyone. Thank you so much again!

I do know one card trick, that you may know, and if I get my internet connection back up and running after my home improvements I’ll make you a video to show you. I can’t wait. Hopefully you won’t know it, but I’m sure you do. Unfortunately I never new it’s name though. I’ll practice it more so it will be as seamless as possible like your videos.

Take care❤

[ShreyRaj]

This concept has already describe by our great mathematician RAMANUJAN

[gagandeep]

*Let three digit number:*
xyz= 100x +10y+ z
*Reverse:*
zyx = 100z + 10y + x
*Subtract:* xyz-zyx = a9b
After subtraction, the result is a9b.. where
Middle digit is always 9 and sum of a and b is always 9.
*Reverse again:* b9a
*Add:*
a9b + b9a = (a+b)/18/(a+b)
We know a+b = 9
Let's put the values:
9/18/9.. Carry 1 of 18 to hundredth place, so it becomes 10/8/9.. i.e. 1089.

No it's not magic.. simple maths...