Sum of the digits of 111111111^2

[@vasiledumitrescu9555]

<a href="#" class="seekto" data-time="265">4:25</a>, bro stared directly into my soul

[@PrimeNewtons]

Lol. I was waiting for the train to pass. I clipped that part out.

[@amritsharmapoudel959]

You should not have clipped that. It was funny and interesting at the same time.

[@jpl569]

Happy New Year to Prime Newtons !

[@PrimeNewtons]

Happy New Year

[@hazevthewolf178]

As a railfan, I love how you patiently wait for the passing train to go by. Yes, rep-unit numbers can be rather interesting. For fun, a little algebra... n(n + 1) / 2 + n(n - 1) / 2 = (n / 2)((n +1) + (n - 1)) = 2n(n / 2) = n^2

[@zanti4132]

When you multiply 111,111,111 by 111,111,111, you know each row will be 111111111, for which the sum of the digits add up to 9, and there will be nine rows. So, 9 × 9 = 81. The only detail to consider in whether the calculation ever requires a carry, but with there being nine rows and with each digit never being larger than 1, it's clear carrying will never come into play. So 81 has to be the answer.

[@jpl569]

Very nice « brute force » explanation !! I love it ! Maybe you will like the « pure algebra » approach : Let X = 111…1 with n digits, then X = ∑ 10^j for 0 ≤ j ≤ n-1. And X^2 = ∑ 10^j 10^k for 0 ≤ j ≤ n-1 and 0 ≤ k ≤ n-1. Just count the number of powers of 10 in X^2 (including 10^0) : there are exactly n^2. This is also the sum of all digits in X^2 ! Thanks for your interesting videos. 😊

[@jpl569]

An other way is to define P (t) = ∑ t^k for 0 ≤ k ≤ n-1, and S (X) as the sum of the digits of X. Then evidently X = P (10), and S (X) = P (1) = n. So : X^2 = P^2 (10), and S (X^2) = P^2 (1) = n^2. And : for all q € N, X^q = P^q (10), and S (X^q) = P^q (1) = n^q. 🙂

[@srisaishravan5512]

Thanks for uploading sir! Your videos are the best

[@sr6424]

Happy New Year from Birmingham England. Last year I put on puzzle evenings in a local bar. Mainly mathematical and logical problems - although I had to put in a couple of word puzzles! Your videos inspired me! You make maths fun! I did it too, but not as well as you! This puzzle would have been great!

[@artandata]

more about the ten one digits square number 111111111^2 * surprisingly the sum of digits is 82 that means only one unit more than 9^2 * number of divisors is 81, one less of its sum of digits * can be factorized as 11^2 × 41^2 × 271^2 × 9091^2 * and of course is a perfect square Have a Nice & Happy New Year ! (for you and the community. regards from Brazil)

[@Wilhelm-mg1jf]

Nice one, it kind of remembered me of Paschal's triangle.

[@Abby-hi4sf]

Happy New year Prime Newton!

[@bjorntorlarsson]

Perhaps not in your taste, but I'd suggest that you do more "mental arithmetics". Going for the primes 7 up. How come the inverse of each prime 7 and larger have a repeating sequence of digitals the length of which is one less than the prime? (Except for 13, unless as if by coincident the other half of the twelve digits happens to be the same as the first half). That the digits of an inverse prime add up to 9's is of course the clue. It has to do with the base ten, doesn't it? 1/7=0.142857... and first half of it plus the second half is 142+857=999. Same for the inverse of ALL primes, if we forget about the very broken figures 2, 3, 5.

[@jay_13875]

Write the number k=111111111 as p(x) = 1+x+x^2+...+x^8 evaluated at x = 10. Note that the sum of digits of k is p(1)=9. Then k^2 is p(x)^2 evaluated at x=10. As long as all the coefficients of the expanded polynomial p(x)^2 are less than 10, we can find the sum of digits of k^2 by evaluating p(1)^2 = 9^2 = 81.

[@AndrewCWSoh]

I would like to see some problems on Vectors, and Matrices. Thanks

[@JossWainwright]

Before watching the video. Doing small cases 11^2, 111^2 it's clear that the multiplication of strings of n 1's gives a staggered stack of strings of 1´s of the length of n that will be added (with zeros at the end of all but the first, by the multiplication algorithm). Adding them will give a symmetrical ascending and descending sequence of digits with the highest digit in the middle. It's like Euler's famous addition of the digits from one to 100, except there are two lists of 1 to n - 1 plus a middle term equal to n. Now the only hitch is if there are carries in the addition, but because the problem has exactly 9 digits, there will be no carries. so the answer is 2*(4*9) + 9 = 81. All strings of n 1's from n = 1 through 111111111 when squared have a digitsum of n^2. At n=10 the pattern is broken

[@kateknowles8055]

Hexadecimal numbers are base 16. You can go up to F digits in hexadecimal before the pattern will break. 11111 11111 11111 ^ 2 's digit sum will be In base 16 : F + 2E+2D+2C +2B +2A + 9 + 51 (incorporating a previous known result, 81 is 51 in hexadecimal) In base 10 : 15 +28 +26 +24 +22+20 +9+81 = 15 +100 +110 = 225 because nothing carried, nothing lost so this is 14 sixteens plus 1 the answer : E1 , is a hexadecimal number

[@AmilQarayev41]

Happy New Year. INTEGRAL.

[@ErickOliveira-i3w]

Happy new year from Brazil, what a fun question!

[@thomazsoares1316]

Achava que eu era o único br que assistia os vídeo dele, mas fala a verdade: ele ensina bem, não?

[@Nature-Melody2106]

Happy New Year, sir ❤🎊🎆

[@ricardoguzman5014]

Recreational math is fun. I thought of a problem a couple months or so ago that is the type that you do on this channel. Find N such that N+1 is a square and 2N+1 is a square. If you're up to it. Happy New Year.

[@wannabeactuary01]

Happy new year too!

[@Kakarot-kr]

It's all about analysis Happy new year from India 💗

[@yajupatel6181]

I got this exact question in a comp test last week and it shows up after I got it wrong :( Happy new year from India ❤

[@AyooTube]

The day 1 of the month 1, is a great day to multiply 1 by 1 like 1111111111111 times... Hahahahah Happy new year, everybody!

[@glorrin]

Great video, But every time I see a cool new trick I want to learn why does it work ? And I believe this one is easy to show, let's say A=1....1 has n+1 digits with 1

[@Ali-i9c2y]

Happy New Year from Indonesia 🇮🇩

[@RyanLewis-Johnson-wq6xs]

Excellent math skills.

[@kragiharp]

Hi! I read a book a long time ago: Vedic Mathematics. There every calculation is checked by doing the exact calculation of the problem with the sum of the digits. You have to get the same answer. So I just count the 1s -> 9. 9^2 = 81. Finished. Although I have never seen a proof that this method always is correct. In case someone knows, please tell me. ❤️🙏

[@HeckYeahRyan]

i remember playing with my calculator and seeing a pattern by doing 11^2 or 1111^2 and i realised each digit will just be counting from 1 to the number of ones and then counting back so 111111111^2=12345678987654321

[@RyanLewis-Johnson-wq6xs]

Sum of digits 111111111^2=81 final answer 111111111^2=12345678987654321 It’s in my head.

[@marcclaude2065]

Fun!

[@maribelle7626]

happy new year 🎉

[@vnms-]

If you go above 9 digits then the sum of digits is 81+(n-9)^2 where n is the number of digits

[@nigerianprinceajani]

111111111² = 111111111 + 1111111110 + 11111111100 + ... + 111111111000000000 = 12345678987654321 so the sum of the digits is 2*8*9/2 + 9 = 9*9 = 81 (using Gauss' sum formula for the sum of 1 + 2 + ... + 7 + 8).

[@RyanLewis-Johnson-wq6xs]

1^2=1 11^2=121 111^2=12321 1111^2=1234321

[@adgf1x]

45+36=81

[@MrJPI]

I did it in my head in less than one minute! Oh yeah! 😅

[@ivanhorbenko7529]

Here comes the beauty of numbers. Nice 😊

[@vinhxnoobx]

Huh i still dont understand the part 2,3,4 and they ^ 😢

[@thomazsoares1316]

Make a video showing the result of sum(n!/nⁿ) for n=1

[@itsphoenixingtime]

i remember that fact about 111,111,111 x 111,111,111 =12345678987654321 Then you can pair all the digits up adding to 9, get 9 9's and that's 81.

[@Dr_piFrog]

Easy ==> 81

[@dbzsongs2474]

81

[@VictirHds]

Sum of digits is 25

[@ishantagarwal1082]

Happy new year, Newton! Btw, did you know that 2025 is the summation of n³ upto n=9? 2025 = 45² = (1+2+3+...+9)²=(1³+2³+3³+...+9³) That means the next time this will occur will be the year 3025, a thousand years from now!

[@mskiptr]

9*9

[@RobG1729]

At least once you say "number" when "digit" is more accurrate. Your videos are so instructive that I check for new lessons nearly every day.

[@DragonX999]

What is the Lim ((sinx)!-1)/x) x->0

[@bjorntorlarsson]

Limb the Sinx, and arrow that ring in the middle of the paper here. My sergeant ChatGPT says when it is drunk at the shooting range.

[@jbasanth4896]

Mm,i tried... As usual , like using the Taylor series or Maclaurian series. Sin(x) =x, when x approaches zero. And since x is small, sin(x) ! Must be small. By using this, Sin(x) ! =0 So, the equation will become, (1-1) /0, the limit does not exist on that point I don't know whether it's a valid proof.

[@DragonX999]

@@jbasanth4896 uhmm ur logic for sinx = x is right but then (sinx)! or (x)! = 0! Which is 1

[@jbasanth4896]

@@DragonX999 oops, i forgot that, ok, i will edit it

[@jbasanth4896]

@@DragonX999 i have another solution, Same, sin(x) =x, The equation will become (x! -1) /x, use L'Hospital's rule, d/dx x! =x!ψ(x), where ψ(x) is digamma function Now, Lim x! ψ(x) =0! ψ(0) x→0 Where ψ0) =-γ which is, -0. 5772

[@priyanshsrivastava3571]

81 , did it without pen , just by a trick of getting the square of given type of numbers