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! :)
@@armlatv2260 that's exactly what I was wondering about. Matter fact I chose 666 when my brother watched this video and tried to do the trick on me. Well now here i am in the comments trying to find what u do it someone pics 111 222 ...
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.
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
@@m1n7o if after subtracting with the reverse number leaves you with two digit than you put 0 For e.g. 152-251= -99 Now you have to put 0 after 99 So it makes 990+099= 1089 Hope I was clear enough 😂😅
That's so awesome..thank you....unless it's not. Ramanujam taxi number(story) is 1729 which is the smallest number which can be expressed as the sum of two different cubes in two different ways.
the number only needs to be not symmetrical. Two of the same are fine as long as they are something like 221 oder 667. Only problem is something like 121 or 444. edit: even 998 is fine as long you treat the result of { 998 - 899 = 99 } as a three digit number of 099. 099 + 990 = 1089
It does not work with all numbers. If you pick 121 the reverse is 121 so you get 0. If you pick 120 the reverse is 012 so you get 108. These are the exceptions for this rule plus when you get 99 as suggested buy others. This rule is not complete.
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
I don't get many free drinks but I do know how to get the cheapest drinks. You see it's this magic power I have, I can drink your drink and then make it appear back in the glass. Don't believe me? Well, I don't wanna take all your money, I'm that confident, so bet you 10c i can do it. *Drink is drunk* *Weird mystic hand gestures and mumbling* Ah well, guess it didn't work, here's your 10c If you do it right whoever you got will start laughing and probably send all his mates round to you as well.
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".
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.
It was sweet and concise until the last part, thanks a lot for the explanation. (but...) (let d = a-c) 99(d)+(inverted_99(d)_) = 99(11) = 1089 I am having a hard time proving that last part but it has to do with prime numbers added to 11. For 2, 2+9 and 9+2, for 3, 3+8 and 8+3, 5 goes with 6, 7 goes with 4. 1 is not a prime number and doesn't go with anything hence i am having a hard time finding the rule that would render the last part of your reasoning slick and concise like the first part was... 1089 is 33 squared, (11)(99), (11)(11)(3)(3). I know the answer is somewhere in this haystack, someone please find my needle lol. Thanks
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
The claim should be that the difference of any number (greater than 9 and not equal to its mirrored version) and its mirrored version, is always divisible by 9. Essentially, it all comes down to this: If a>0; x, y ≥ 0; x>y and a, x, y are all integers: D = a * (10 ^ x) - a * (10 ^ y) = = a * ((10 ^ x) - (10 ^ y)) = a * M D is divisible by 9 if M is also divisible by 9. M is always divisible by 9 for the same reason the differences 1000-10=990 and 100-1=99 are all divisible by 9.
*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...
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.
To everyone! Every 3 different digits works if it's done in the right way! 🙃 Edit To replies:- I can't calculate every number so here are the two common mistakes- 1) Listen carefully in the video if you get an answer in negative ➖ remove that and make it positive! 2) If you get an answer as a two digit number (most commonly 99) add 0 before the number so 99 becomes 099 then add 099+990 = 1089 🤓
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❤
I have written a program to check all the 3 digits numbers. Only 640 have 1089 answers. The rest is incorrect because it will encounter a palindrome during the process.
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
If they get a negative number after the first step, ignoring it is a bad idea, because if they don't clear it out and re-type it correctly as a positive number, this will not work.
Look it in this way... Take any 3 digit number and calculate it's difference from it's opposite number. Use word difference. Difference of any two numbers is considered positive.
No need to clear it out. Simply multiply by "-1", OR subtract its reverse instead of adding it, OR click on the change sign (+/-) function (if one is available).
When I first heard this, I was told no repeat numbers. But it doesn't work if you say: 132 or 796 , 495, 738, etc. If the 1st and 3rd numbers are sequential, the inverce makes 99.
I didnt knew such kind people existed...i have been watching it lately and i like it...tbh those 5 to 10 min magic vid do not teach me magic as much your shorts does!!your smile is too decent
@@harshitgaming9962 The reverse of 2002 is itself, and not a different number (with the emphasis on ‘a different number’); hence it doesn’t count. Any number could divide itself, so what! We are talking about the nontrivial reverse divisors. Palindromic numbers have reflectional symmetry across the vertical axis.
The phenomenon you're describing with the number 1089 is a well-known mathematical trick. It works with most three-digit numbers, as long as the first and last digits differ by at least two. Here's how it works: Choose any three-digit number where the digits are not all the same (like 651). Reverse the digits to form a new number (156 from 651). Subtract the smaller number from the larger number (651 - 156 = 495). Reverse the digits of this result (594 from 495). Add this number to the original difference (495 + 594). The final result is almost always 1089. This trick works because of the properties of number manipulation and digit rearrangement
90s baby here. When the gameboy got boring i used to play with a calculator and discovered this when i was a wee kid. I thought i was a genius xD showed everyone in school the next day
It does work... So after subtraction, the resultant number will be 99, so add the reverse of it which is 990 (since 99= 099 so the reverse should be 990) If you find a two digit number write a zero infront of it, which is 099, Add 099+990 = 1089
You can technically do it with larger than 3 digit numbers, but you need to start adding in a decimal place to make it equal 1089. Otherwise you start getting things like 108090.
@@USE_YOUR_BRAIN So, I realize it's not how you would probably normally treat a decimal. But yeah, if you use ###.# instead of ####, it will work. Last I tried at least.
Let the unique digits of the number be *x* , *y* , *z* in order. So the number is: *100x + 10y + z* And the reverse of it is: *100z + 10y + x* Subtracting the reverse from original gives: *99(x-z)* If *x>z* let *t=x-z* , else let *t=z-x* So the number we have, ignoring the sign, is: *99(t) = 100(t) - t* *= 100(t-1+1) - t* *= 100(t-1) + 100 - t* *= 100(t-1) + 10(9) +(10 - t)* Possible values of *t* : 1 to 9 *(t-1)* : 0 to 8 *(10 - t)* : 1 to 9 so they are single digits. So the number we have is: *100(p) + 10(9) + q* where *p = (t-1)* *q = (10-t)* The reverse of this number is: *100q + 10(9) + p* Adding these together, we have: *101(p + q) + 180* *= 101(t-1 + 10-t) + 180* *= 101(9) + 180* *= 1089*
cause you arre not respecting the request of the magician: "numbers have to be different". So you can't choose numbers like 111 or 181 or 226, cause they are not all different numbers
@@usernameusernamwell it works for the numbers you mentioned. It is just that you ignore the negative sign if the difference is negative and if the difference comes out as a 2 digit number you should consider it as 3 digits with 100's place being "0" and proceed. For example for 102 102-201=-99 Ignore the -ve sign and consider it as 099 099+990=1089
One thing she might forgot to say that "the first 3 digit number must be made of non identical numbers" Suppose if anybody takes 666 as his/her 3 digit number in mind then the trick will be failed at the very first step😊
@@pucheou483-384=99 We know 99 can be also written as 099 which is the same thing Now 990+99=1089 See if your results ends with two digit number for example:99 then add a zero in front of it.
@pucheou You absolutely can. Let me explain how, 483-384 is 99 right? Well guess what, because the result is only 2 numbers, the numbers have to be 3 numbers at all times, so what you do in this case is force a 0 at the end in each result of 2 numbers, which is the default number that you start with, 990+099? 1089, but in the calculator you can't put 0 in the beginning so you have to do 990+99, and there you have it
This is similar to the ones that come out negative. Add Zero in the start of it, like 99 would turn to 099, then add 990, that would equal 1,089. Also the ones that don't work are similar numbers like 111, 222 and so on, that would equal to Zero, that's why the request is Different numbers.
The mystery is this: first, take any three digit number, where the first and last digits differ by two or more and reverse the number to produce a new one. Then subtract the smaller from the larger producing another new number. If you reverse this number and this time add the two, the result will always be 1089. For example, if we begin with 452, reversed we get 254, and taking the smaller from the larger, 452 - 254 = 198. If we reverse this number and then add the two we get 891 + 198 = 1089. And it always will be, regardless of the three digit number you start with. To mathematicians of course there is no mystery and an explanation is given below. Let us assume that the initial number is the larger and has digits a, b and c. So, when we reverse and subtract we will have (100a + 10b + c) - (100c + 10b + a) This is the same as 100a + 10b + c - 100c - 10b - a = 99a - 99c = 99(a - c) Because a and c are whole numbers, at the end of the first part of the process we will always end up with a multiple of 99. The three digit multiples of 99 are: 198, 297, 396, 495, 594, 693, 792 and 891. Now, note that the first and last digits of each number add up to 9. So, when we reverse any of these numbers and add them together we get 9 lots of 100 from the first digit, 9 lots of 1 from the third and two lots of 90 from the second and so we get 900 + 9 + 180 = 1089. No mystery at all really!
