Squares Ending in Repeating Digits

[@blackpenredpen]

Super interesting! Your content is always interesting and unique! 👏

[@Gojo_12371]

I have found you to comment on his several videos ✨

[@arjuntilak5023]

Wow, u also watch this....

[@DrBarker]

The ultimate seal of approval - thank you!

[@joaozin003]

Your content is also super unique!

[@elephantdinosaur2284]

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.

[@DrBarker]

This is very cool - thank you for sharing!

[@perappelgren948]

Seeing this over and over again. So simplistic, yet elegant and mindblowing.

[@sweetcornwhiskey]

Does this mean therefore that 38 is the number whose square has the largest number of nonzero repeating digits at the end?

[@englematics]

I think you're right. Good observation.

[@coragon42]

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.

[@radadadadee]

excluding zero. (100)^2 is 10000

[@heygooooooooo]

Loved the fade to "erase" the board.

[@hvok99]

This was really fun! Thanks for taking us on so many guided tours of interesting math. Your channel is gold 🐳

[@DrBarker]

Thank you!

[@tamirerez2547]

Can't explain why, but this video is like a fascinating action movie with a ... SAD END.

[@paradoxica424]

corollary: there is no square root of …44444444 in the 10-adics

[@abdullahh283]

The videos are really great, thanks for the uploads doc!

[@vijaysubramanian2037]

This is cool!

[@emanuellandeholm5657]

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-facedclown]

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-facedclown]

(insert question mark here)

[@shruggzdastr8-facedclown]

Another question that occurred to me is: What if you switch bases (like base-12, for example)?

[@oledakaajel]

An integer that ends with five repeating digits or more also ends with four repeating digits

[@alvaroyonekura455]

Wooow it's a really nice problem !! Very intresting and beautiful solution :)

[@sepdronseptadron]

what about for a different base of numbers?

[@DrBarker]

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.

[@ScorpioneOrzion]

@@DrBarker What about all bases up to X?

[@kenhaley4]

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?

[@tsawy6]

How many solutions are there with three repeating digits?

[@DrBarker]

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_hewillow]

That is an absolutely amusing video!

[@sdspivey]

What about 5 repeating fours, or 6, or 7...?

[@LinkAranGalacticHero]

He proved at the end of the video that it's not possible

[@DrBarker]

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.

[@tomcummings2436]

This is an Irish math Olympiad 1989 problem

[@dp121273]

Cool video but too heavily compressed audio.

[@menjolno]

You forgot 10 sacred. Lol