Super interesting! Your content is always interesting and unique! 👏
@Gojo_123712 жыл бұрын
I have found you to comment on his several videos ✨
@arjuntilak50232 жыл бұрын
Wow, u also watch this....
@DrBarker2 жыл бұрын
The ultimate seal of approval - thank you!
@joaozin0032 жыл бұрын
Your content is also super unique!
@elephantdinosaur22842 жыл бұрын
Nice problem and solution. If you look at the same problem but for cubes there's infinitely many solutions. As an example the cube of 79,858,716,637,368,288,471 ends with 20 ones. We'll prove something a little stronger which will make the proof by induction easier. We'll prove the claim that for every k we can find an integer x such that x^3 = 111...111 mod 10^k and x^2 = 1 mod 10. The base case k = 1 follows by considering x = 1. For the inductive step we'll suppose we can find an x such that the assumptions are satisfied for some k. We'll show we can find an X that satisfies the same equation but with k+1. We want to show X^3 = 111...111 = (10 ^ (k+1) - 1) / 9 = - 1 / 9 mod 10 ^ (k+1), or 3 + 27 X^3 = 0 mod 10^(k+1). As well as X^2 = 1 mod 10. Substituting X = x + 10^k * u, we want to solve the below equation in u: 0 = 3 + 27 * X^3 = 3 + 27 * [x^3 + 3 * x^2 * 10^k * u + 3 * x * 10^(2k) * u^2 + 10^(3k) * u^3] = [3 + 27 * x^3] + [81 * x^2] * 10 ^ k * u mod 10 ^ (k + 1) By the inductive hypothesis we have: x^2 = 1 mod 10 so in particular 81 * x^2 = 10B + 1 for some B and X^2 = (x + 10^k * u)^2 = x^2 = 1 mod 10. Additionally we have 3 + 27 * x^3 = 0 mod 10^k, i.e. 3 + 27 * x^3 = A * 10^k Substituting the equations for A & B in gives: A * 10^k + (10B + 1) * 10 ^ k * u = 0 mod 10 ^ (k + 1) A * 10^k + 10 ^ k * u = 0 mod 10 ^ (k + 1) A + u = 0 mod 10 u = (- A) mod 10 Thus we've found our u and we're done.
@DrBarker2 жыл бұрын
This is very cool - thank you for sharing!
@perappelgren948 Жыл бұрын
Seeing this over and over again. So simplistic, yet elegant and mindblowing.
@sweetcornwhiskey2 жыл бұрын
Does this mean therefore that 38 is the number whose square has the largest number of nonzero repeating digits at the end?
@englematics2 жыл бұрын
I think you're right. Good observation.
@coragon422 жыл бұрын
Actually, there is a repeating pattern. 38+500n and 462+500n (where n is an integer) will always have perfect squares ending with 444. I used the fact that 12+50n and 38+50n always have perfect squares ending in 44 to test this, but these sorts of patterns can be proved using binomials.
@radadadadee2 жыл бұрын
excluding zero. (100)^2 is 10000
@heygooooooooo2 жыл бұрын
Loved the fade to "erase" the board.
@hvok992 жыл бұрын
This was really fun! Thanks for taking us on so many guided tours of interesting math. Your channel is gold 🐳
@DrBarker2 жыл бұрын
Thank you!
@tamirerez25472 жыл бұрын
Can't explain why, but this video is like a fascinating action movie with a ... SAD END.
@paradoxica4242 жыл бұрын
corollary: there is no square root of …44444444 in the 10-adics
@abdullahh2832 жыл бұрын
The videos are really great, thanks for the uploads doc!
@vijaysubramanian20372 жыл бұрын
This is cool!
@emanuellandeholm56572 жыл бұрын
This is cool! I was thinking about k^2 = y 10^m + x ( 10^m - 1 ) / 9, using a finite geometric series. Using difference of squares, this gives (3k)^2 - (3 sqrt(y) sqrt(10^m)^2 = x * 1....111 => (3k - 3 sqrt(y) 10^(m/2)) (3k + 3 sqrt(y) 10^(m/2)) = x.xxx, which gives a nice factorisation if y is a perfect square and m is an even number.
@shruggzdastr8-facedclown2 жыл бұрын
The obvious question for me is: Given that we've proven that it's impossible to have the square of an integer contain four repeating end-digits, how about checking for repeating digits of five or more, and then see if there's a pattern with respect to which one's work and which ones don't or if there is no pattern and three repeating end-digits is the maximum.
@shruggzdastr8-facedclown2 жыл бұрын
(insert question mark here)
@shruggzdastr8-facedclown2 жыл бұрын
Another question that occurred to me is: What if you switch bases (like base-12, for example)?
@oledakaajel2 жыл бұрын
An integer that ends with five repeating digits or more also ends with four repeating digits
@alvaroyonekura4552 жыл бұрын
Wooow it's a really nice problem !! Very intresting and beautiful solution :)
@sepdronseptadron2 жыл бұрын
what about for a different base of numbers?
@DrBarker2 жыл бұрын
This is a really interesting question! In binary, you can show that all squares have to end in ...01 or ...00, so there are no nice squares ending in repeating non-zero digits. I'll have to think some more about all the other bases though.
@ScorpioneOrzion2 жыл бұрын
@@DrBarker What about all bases up to X?
@kenhaley42 жыл бұрын
Nicely done! As are all your videos. I'm sure I'm not alone when I ask...can you give us a little personal background? Like, what is your full name, how old are you, and what is your educational background?
@tsawy62 жыл бұрын
How many solutions are there with three repeating digits?
@DrBarker2 жыл бұрын
There are infinitely squares ending in 3 repeating digits. To see this, you could consider 1038^2, 10038^2, 100038^2, ... , where adding more zeroes doesn't affect the last 3 digits. But there are other squares ending in 3 repeating digits too (e.g. 462^2, 538^2, and more).
@matt_hewillow2 жыл бұрын
That is an absolutely amusing video!
@sdspivey2 жыл бұрын
What about 5 repeating fours, or 6, or 7...?
@LinkAranGalacticHero2 жыл бұрын
He proved at the end of the video that it's not possible
@DrBarker2 жыл бұрын
Yes, if there were 5 repeating fours at the end of a number, then the last 4 digits would also all be fours. But we've seen that that is not possible, so you can't have any more than 3 repeating fours at the end.