The most creative digit sum problem I have ever seen!!

[@Buridan84]

8:21 that flip and editing are so smooth

[@Didnddixjxhjxxj]

I really missed the backflips

[@Maths_3.1415]

8:19

[@anapatriciagrimolizzi9560]

It's excepcional ​

[@colbyforfun8028]

We can add the digits of the numbers in any order we want. Including leading zeros, in the set of all n-digit numbers each digit appears equally often, or 1/10th of the time. There are 10^n n-digit numbers and each has n digits, so there are n*10^n digits in total and each of the 10 distinct digits occurs n*10^(n-1) times. Summing all this brings us n*10^(n-1)*45 where 45 is the sum of a single set of each digit.

[@allozovsky]

Neat solution 👍

[@ingiford175]

This is what I was thinking...

[@jaimeduncan6167]

Great

[@GrouchierThanThou]

You spilled the beans! The 9th triangular number is supposed to be a secret that only Simon's very best friends get to know.

[@ingiford175]

That would be an interesting colab....

[@lisandro73]

8:23 literally jumping to the prove

[@goodplacetostop2973]

15:24 Good Place to Sssss 🐍

[@JOSHUVASRINATH]

🐍🐍🐍

[@plislegalineu3005]

tbh he is a vampire born in the 1700s

[@MathFromAlphaToOmega]

There's a really interesting problem like this, but with the product of the nonzero digits instead. If you write p(n) as the product of the nonzero digits of n, then p(1)+p(2)+...+p(9)+p(10)=46, p(1)+p(2)+...+p(99)+p(100)=46^2, and so on. A quick way to prove it is to expand (1+1+2+3+4+...+9)^n and show that it gives exactly p(1)+p(2)+...+p(10^n).

[@Maths_3.1415]

I think I have seen this problem on AOPS. 🤔

[@MathFromAlphaToOmega]

@@Maths_3.1415 Yeah, I think it's from a math contest, but I can't remember which one.

[@m9l0m6nmelkior7]

8:20 why would you prove that by induction ??? there is a logical way of finding that… without dropping the formula out of nowhere ! It's quite clear when you see it : When you manipulate all numbers with n or less digits, you have in hand all possible combinations of n digits !, so in total you do have 10^n numbers, but since each number is n digits long (counting 0s as digits in say 0001), you do get n*10^n digits in total ! but each one of them has exactly 1/10 chance of being any specific digit between 0 and 9, so every digit appears n*10^n /10 times so to have the sum of all those digits, you simply have to sum numbers from 0 to 9 (9*10/2 = 45) and then multiply by that n*10^(n-1) ! You can just find this an = 45 n 10^(n-1) formula in quite a simple way !

[@CTJ2619]

jumping into our proof .. then does a backflip - a great video

[@Alan-zf2tt]

It is amazing really - not just the math I also include Michael's enthusiasm at sharing insights and wonder = brilliant!

[@charlesbrowne9590]

There is a much easier route to the solution. Just realize that ΣS(n) from n=0 tp 10^M -1 is just the sum of all digits present. There are 10^M such numbers with M decimal places. One tenth of the places goes to each digit 0…9. So the sum is 45*M*10^(M-1).

[@KitagumaIgen]

Nice to have the backflips back! But I miss the blackboard-erasing knuckle-tap even more...

[@jursamaj]

Hmm. I didn't break it into the difference of 2 sums, nor did I go the induction route, but just figured the number directly: A=2003 (the exponent in the lower limit of the sum). The total number of numbers being summed is the difference in the upper and lower limits of the sum, plus 1: 10^(A+1)-1-10^A+1 = 9*10^A. The 1st column is equally split between the digits 1 thru 9, so it's average digit is 45/9. Multiply by the number of numbers: (45/9)*9*10^A = 45*10^A. Every other column is equally split between the digits 0 thru 9, so it's average digit is 45/10. Multiply by the number of numbers: (45/10)*9*10^A = 405*10^(A-1). There are 1 '1st column' + A 'other columns': 45*10^A+A*405*10^(A-1) = 450*10^(A-1)+A*405*10^(A-1) = (450+A*405)*10^(A-1). Since A=2003, the result is (450+2003*405)*10^2002 = 811665*10^2002. Sum of digits is 27. As expected, using the above formula for different values of A always yields multiples of 9. But those multiples bounce around in a way I can't immediately spot a pattern in.

[@shahzadnaeem3717]

Nice backflip! 🙃

[@speedbird8326]

@8:21 And that's a good place to keep replaying...

[@fortetwomusic]

The flip IS the induction proof!

[@matthewleadbetter5580]

That's a good place to st-

[@wojciechwisniewski6180]

Good old Michael, backflip is back! ;)

[@mikenielsen8781]

Very nice. I may have had math lecturers as good as you are all those years ago, but if I did, I was too dumb to know it. But I am pretty sure they never did backflips.

[@Flukeworm]

(Request) Find all natural numbers a and b ,such that 5^a +2^b +8 is a perfect square.

[@HenkVanLeeuwen-i2o]

He is amazing.

[@fatbudyn]

okeeey the backflip was fire

[@mskellyrlv]

No wonder I have such difficulty with induction. I could never do a backflip like that, even as a kid...and I'm almost 70.

[@romanbobyor]

jump????

[@mnokeee]

cool backflip

[@usercy3pd5lj2i]

GF: I'm pregnant! Me : 15:22

[@charleyhoward4594]

this s(n) ISNT the sum of the first n natural numbers (n (n + 1)) / 2 ??

[@bjornfeuerbacher5514]

No, it isn't. It's the sum of the digits of n. As is stated right there on the board, and as is said explicitly by Michael.

[@pwmiles56]

Phew, I did this without notation and eventually got the same answer. Writing out the numbers in the sum they are 1000... 000 1 followed by 2003 0's 1000... 001 1000... 002 . 9999... 999 2004 9's There are 10^2004 - 10^2003 entries overall i.e. 9x10^2003. In all but the first column we have 0,1,2,.. 9 occurring with equal frequency, mean 4.5. In the first column we have 1,2,... 9, occurring with equal frequency, mean 5 Therefore N = 9 x 10^2003 x (2003 x 4.5 + 5) = 9 x (2003 x 45 + 50) x10^2002 = 811665 followed by 2002 0's s(N) = 27

[@blairhoughton7918]

This. Adding up all the numbers from 0, then subtracting the sum of all numbers shorter than 2004 digits, seems like extra work.

[@keithmasumoto9698]

Very nice! Thank you

[@raneynickel1]

Mr.micheal Penn u suggest any book for high school advance discrete mathematics book ?

[@JacobPlat]

(10^2004) -1 is just 2003 9's, or not?

[@JacobPlat]

I mean 2004 9's on a row.

[@jaimeduncan6167]

Súper creative the problem

[@El0melette]

8:20 xd

[@cycklist]

Austrian-Polish? How weird. They don't even border each other.

[@alipourzand6499]

I was expecting 42 but 27 is fine ☺ great video as usual

[@blairhoughton7918]

Seems like a long way to go, calculating all the way up from 1+digit numbers, instead of just figuring out a formula for the sum of all the 2004-digit numbers, which is what the original sigma over k appears to be.

[@colinn4239]

hey i was actually able to solve this one!

[@ezequielangelucci1263]

and what about the single digit from the 10 to the 2002 power? I think the answer should be a little higher

[@bernardlemaitre4701]

❤

[@piszczuch3374]

polska gurom

[@VideoFusco]

We need to calculate the sum of the digits of the numbers from 10...0 (a "1" followed by 2003 digits "0") to 9...9 (2004 digits "9"). In total we have 9*10^2003 numbers. In all positions from one to place 2003, all the different ten digits appear with the same frequency, while in place 2004 only the digits "1" to "9" appear The sum of a single sequence of digits from "0" to "9" is (9+0)*(10/2)=45. For each of the seat positions less than or equal to 2003 each digit is repeated 9*10^2003/10=9*10^2002 times, therefore the sum of the digits for all these positions is 2003*45*9*10^2002= 811215*10^2002. In position 2004 each digit repeats 9*10^2003/9=10^2003 times, so the sum of the digits for this position is 45*10^2003=450*10^2002. The sum of all the digits is therefore 811215*10^2002+450*10^2002=811665*10^2002.