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 ...
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
@@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.
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
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".
Such a delight sitting here here india and want to learn a little 🎉 fun skill to pass your my only way to learn really liked your videos and my favourige vedio of uours is sybic cut which was done 7 years ago !!!!!!!!! But nonethless you look toooo beautiful thank you man😅❤
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.
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.
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
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.
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
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.
*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...
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
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❤
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
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 🤓
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.
Also if you do the first part where the subtraction happens, you can predict the result of the operation if you know the first or last digit of it. middle digit is always 9 and the sum of the other two digits is always 9. So in the example of the video we have 495, knowing 5 you can find 4 and construct 495. People will be surprised. The only case where this doesn't work is with 3 digit numbers where all digits are the same (eg 444, 888 etc)
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).
If you take 756 and subtract it reverse from original 756-657=99 Now adding reverse of 99 which is 99 we get 198 Edit:I found a new thing that if the last and first digits have a difference of 1 then after subtracting the reverse of the number from the original we get 99 and adding reverse of 99 which is 99 we get 198 👍
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*
3 different numbers even don't need. You can assume atleast 2 same numbers either on left side of 3 digit set or at right side of them, for example 221 or 122 or 488 or 884. These sets of numbers also work.
Below is a very very long and rather convoluted mathematical explanation of this problem. Let the first digit of the first number be a, second digit be b and third digit be c. Then, first number = 100a + 10b + c Reverse of this number= 100c + 10b + a First number - its reverse = 100a + 10b + c - (100c + 10b + a) = 99a - 99c = 99 (a - c) The next set of calculations is to find the 3-digit formula of the above number. = 99 (a - c - 1) + 99 = 100 (a - c - 1) - a + c + 1 + 90 + 9 = 100 ((a - c) - 1) + 90 + (10 - (a - c)) Thus, we have the first digit of the difference as (a - c) - 1, second digit as 9, and third digit as 10 - (a - c). The reverse of the above number is = 100 (10 - (a - c)) + 90 + ((a - c) - 1) Sum of the above 2 numbers = 100 ((a - c) - 1) + 90 + (10 - (a - c)) + 100 (10 - (a - c)) + 90 + ((a - c) - 1) = 100(a - c) - 100 + 90 + 10 - (a - c) + 1000 - 100(a - c) + 90 + (a - c) - 1 = 1000 + 90 - 1 = 1089 One peculiar scenario where this rule might fail is when the difference of the first number and its reverse is 99. E.g. 201 - 102 = 99 In this case, 99 + 99 = 198 which is not equal to 1089. However if we consider 99 as a 3-digit number, that is 099, then we have 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.
She's a woman, she use Google pixel. ... I'm so proud of her. This comment is just short comment . But it already has 50LIKES, and it will end at 82 LIKES. L😄L.
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😊
