PyMath #4 - The Curious 2019 AIME 1 Problem 1 Exercise!

[@NPCooking69]

*Make sure to check out the providers of the best Christmas present this year! :D **brilliant.org/FlammableMaths* Also, Christmas Merch :00000000 teespring.com/new-merry-christmath-golden-ra

[@misterg3tr3kt11]

how is this comment 4 hours old? 4 hours < 4 seconds?🤔🤔🤔

[@angelmendez-rivera351]

@@misterg3tr3kt11 The video could've been unlisted.

[@misterg3tr3kt11]

@@angelmendez-rivera351 yeah

[@latefoolstalk676]

Loving those nines that look like "g" because everyone knows π2 = 9 = g

[@oferzilberman5049]

If we say π=3 Then π²=9=g

[@michaelempeigne3519]

pi squared does not = g

[@michaelempeigne3519]

( 16*arctan( 1 / 5 ) - 4*arctan(1 / 239 ) )^2 is not equal to g

[@minh9545]

@@michaelempeigne3519 is this a joke? If it is not, you are clearly missing the point here.

[@abcxyz4207]

Im so sick of hearing this

[@OlliWilkman]

The main way to improve the Python code would be to loop over the characters of the string ("for digit in x"), that way you don't need to make a list nor use the range function. The most "pythonic" way to write that function would probably be the one-liner "return sum(int(digit) for digit in x)", which is short and also easy to see what it does immediately.

[@OlliWilkman]

Another trick to help build big number strings like that is that you can multiply a string by an integer, so 321*"9" gives you a string of 321 nines.

[@АлександрКовалев-о7я]

Me: "Monkey Panic"

[@АлександрКовалев-о7я]

@@OlliWilkman kzbin.info/www/bejne/rmSpg6Stgqtpaq8 I wish you a happy watching

[@mortadhaalaa5907]

print(sum( int(d) for d in str(sum( int(k*'9') for k in range(1, 322)) ) ))

[@gudmundurjonsson4357]

nice

[@a-rod6336]

You don't actually need to write range(1,322), range(322) does that by default

[@mortadhaalaa5907]

@@a-rod6336 if you run it, it gives an error because range(322) starts at zero and int(0*'9') is undefined

[@a-rod6336]

@@mortadhaalaa5907 that's right lol I suppose you would need to add 1 to k

[@angelmendez-rivera351]

Note 2: Let a = floor[ln(n)/ln(10)] + 1, and let 0 < n < m. If 10^a | m, then digitsum(n + m) = digitsum(n) + digitsum(m). However, in the video, this was not possible to use, because -321 < 0.

[@adonisnunez6078]

Here is my solution by using map() and a lambda funtion: suma = sum(list(map(lambda x: 10**x, list(range(1, 322))))) - 321 suma_digitos = 0 for d in str(suma): suma_digitos += int(d) print(suma_digitos) Greetings from Dominican Rupublic, I really like this chanel.

[@gudmundurjonsson4357]

Tried optimising my own code with the tips from Olli Wilkman's comment: def digisum(x): s = sum([int(digit) for digit in str(x)]) return s number = 0 for i in range(1,321+1): number += 10**i - 1 print(digisum(number))

[@OlliWilkman]

The square brackets around the for expression aren’t necessary. They create a list which you never use again and the sum function works with just the generator expression. Also why assign it to a variable rather than just returning it?

[@OlliWilkman]

You could also write that second part as “number = sum(10**(i+1) for i in range(321))” if you want to be terse. If your loop is just doing one addition to an accumulator variable, you can usually replace it with a sum function with a loop expression.

[@OlliWilkman]

Personally I’d create the number with number = int(321 * ”9”), which is a slightly cheeky way I feel.

[@paulmoat7047]

slightly smaller code def digsum(x): return sum(map(int, list(str(x)))) print(digsum(sum((10**(i+1)-1) for i in range(321))))

[@copperfield42]

13:17 flammy, you can iterate over the list itself, no need for this range thing (tho technically correct, is ugly as sin), or even over characters on the string, and a string is kind like a list of characters, so like this: def digsum(x): dig = 0 a = list(x) for n in a: dig += int(n) return dig but if you can call list on x, that mean you can iterate over it, so put it directly in the for loop: def digsum(x): dig = 0 for n in x: dig += int(n) return dig and that is the long way, the short way is using list comprehension method along side the sum function def digsum(x): return sum( int(n) for n in x ) and to eliminate the need to call str into your input beforehand, you can include into the function itself, in either of the versions def digsum(x): dig = 0 for n in str(x): dig += int(n) return dig def digsum(x): return sum( int(n) for n in str(x) )

[@PapaFlammy69]

Thank you David!!

[@DatKid321]

Gave my own coding skills a go def digitsum(x): return(sum([int(digit) for digit in str(x)])) n = len(str(321)) print(321 - n + (n*10 - digitsum(321)))

[@monkerud2108]

i played you on lichess, big jens. this channel is still great

[@angelmendez-rivera351]

Consider N(n) = 9 + 99 + ••• + 9•••9, with the last summand having n digits, all digits being 9. Hence N(n) = (10^1 - 1) + ••• + (10^n - 1) = 1•••10 - n, where 1•••10 has n + 1 digits, with n digits being 1. Now, this is equal to 1•••10•••0 + (1•••10 - n), where 1•••10•••0 has x digits being 1, and y digits being 0, with x + y = n + 1, and 1•••10 has y - 1 digits being 1, with y - 1 being equal to the number of digits of n. Therefore, digitsum[N(n)] = n - floor[ln(n)/ln(10)] - 1 + digitsum(1•••10 - n). 1•••10 can be written as 10 + ••• + 10^(floor[ln(n)/ln(10)] + 1), since floor[ln(n)/ln(10)] + 1 is the number of digits of n. This is also equal to 10·(1 + ••• + 10^floor[ln(n)/ln(10)])= [10^(floor[ln(n)/ln(10)] + 2) - 10]/9. Therefore, digitsum[N(n)] = n - floor[ln(n)/ln(10)] - 1 + digitsum([10^(floor[ln(n)/ln(10)] + 2) - 10]/9 - n). This is the closed form for digitsum[N(n)]. In the special case of n = 321, the results of this video arise.

[@angelmendez-rivera351]

Note: the digital root of a number is not the same as the digit sum of a number. The digital root of N is 9, trivially, because the digital root is defined by iterating the digit sum function as many times as necessary until the output is a single-digit number. However, the digit sum merely adds the digits.

[@Skrownia]

I see some people have mentioned the sum function, but I would also like to recommend looking into the map function it will make you able to calculate the digit sum like so "sum(map(int, x))" where x is your number as a string.

[@niki3352]

the digsum function can be a oneliner: def digsum(x): return sum(int(y) for y in x)

[@gudmundurjonsson4357]

think you need int(x) but yeah, thats the nices way to do it i think

[@TheMazyProduction]

The whole program can be one line of code

[@gudmundurjonsson4357]

@@TheMazyProduction doesnt look that nice tho

[@maypiatt3766]

I think this is an AIME problem!! Except there just weren’t as many terms

[@PapaFlammy69]

Oh really? :0

[@maypiatt3766]

@@PapaFlammy69 this is the AIME I 2019 problem 1

[@maypiatt3766]

@@PapaFlammy69 oh and it actually was 321 terms too

[@rupachakraborty3681]

Great video. Btw, which IDE were you using?(You are using black theme. I see, you are a man of culture as well.)

[@dizmat.mp3]

Bruh lmao is that an anime mousepad

[@HAL-oj4jb]

Seems you're new to the channel lol

[@zvxcvxcz]

Python actually is appropriate here because of arbitrary length integers... It may be less trivial to code in other languages.

[@path_selector]

great video. combines 2 of my favorite things! math and programming :)

[@PapaFlammy69]

:)

[@AlexandrBorschchev]

you should create a recursive function too for fun

[@dado070998]

Yo papa what is that figure on your desk? Looks epic.

[@dozzco2827]

God I've been missing these videos!, Brilliant Python Math Problem! 👍

[@devincustodio2258]

Hey have you thought about a recursive implementation for digsum? That would be kind of cool to write

[@copperfield42]

you want it recursive? yeah sure, that is easy here you go: def digsum_r(n): if 0

[@devincustodio2258]

@@copperfield42 that's what I wrote too

[@bbblaesing]

I posted a really stupid response earlier and realized I was summing 9^n for the sum of 1 to 223. I fixed it: def summer(n,g=9): gs=0 for i in range(n): nummer=0 for j in range(i+1): nummer += g*10**j gs += nummer return gs

[@bbblaesing]

answer = summer(223)

[@robertstorlind2302]

And we can use the theorem of the digit sum being congruent to the number mod 9 as a sanity check of the answer 342, right?

[@HAL-oj4jb]

PyCharm is really a charm, my favourite IDE

[@korigamik]

Try thonny

[@thaihm]

Loved the lecture! 👍🏽

[@mathssolverpoint6059]

I solved this question in my head in 2 minutes

[@Pneuma7]

just what I needed, love u papa

[@maxwellsequation4887]

Christmas = SIR ISAAC NEWTON's birthday Ophence

[@bisheshadhikari6508]

Awesome video!! Really enjoyed it.

[@devinwoods641]

If I’m looking to start learning more beyond highschool math where should I start with your videos?

[@akshat9282]

isn't a digit dum supposed to be a single digit aur recursively doing the process on the number. I'm probably wrong but that's what I heard. calculating that would be really easy. take 9 common from the original number. you'll be left with (1+11+...1...1). since you took 9 common, the number is a multiple of 9 and so the digit sum is 9. also, you can use a very basic python one liner for what you did. this wasn't a v nice program daddy but still appreciated ❤️

[@angelmendez-rivera351]

No, that is a different function. What you are referencing is the digital root function, which is not the same as the digit sum function.

[@Evan-ne5bu]

Man wtf, I've never been so early

[@michaelempeigne3519]

sum of digits of a multiple of 9, will yield a digital sum of 9; so wouldn't it make sense that it be 9 regardless of length ?

[@angelmendez-rivera351]

The statement you wrote is incorrect. 99 is a multiple of 9, because 99 = 9·11, but digitsum(99) = 9 + 9 = 18, not 9. In this video, digitsum(N) = 342, not 9.

[@michaelempeigne3519]

@@angelmendez-rivera351 but the digital root of those numbers are 9

[@angelmendez-rivera351]

@@michaelempeigne3519 The digitsum function and the digital root function are not the same. The digit root of these numbers is 9, but not the digit sum. So you are still incorrect per my comment.

[@mehulborad2400]

10:09 Programmers be like : it and IDE not a EDITOR for God's sake

[@chicken2896]

where did you buy your blackboard and for how much?

[@PapaFlammy69]

some 1.5k from some german manufacturer ^^'

[@chicken2896]

@@PapaFlammy69 Thanks 🐐

[@roeesi-personal]

I know this video is old but one liners are always nice so I want to share mine with everyone who will see this comment. sum(int(d) for d in str(sum(int('9'*j) for j in range(1, 322)))) It's not very efficient, but it's a simple python one liner, what did you expect?

[@gapdragon01]

"Three hundred twenty FIRST power"!!!!!!

[@PapaFlammy69]

:^D

[@Chr0nalis]

Try out the visual studio code dark plus colorscheme for Pycharm. Nicest looking one that I've used so far. Darkula sucks balls.

[@raiymshuak]

I think I saw this in the AIME one of the last years

[@matron9936]

Hey saw ya doing it in discord, nice lemme watch dat video see if I solved it right.

[@a_llama]

I saw this problem on mu prime math!

[@alwysrite]

please post more python videos

[@69erthx1138]

If this vid doesn't PROVE that arithmetic is the most difficult kind of mathematics, I don't know what does?

[@angelmendez-rivera351]

...it doesn't prove that. This exercise was surprisingly easy to solve.

[@69erthx1138]

@@angelmendez-rivera351 My post was a joke😁.

[@brettstafford9665]

I solved this in like 3 minutes.

[@ricardoparada5375]

Relatable meme lmao

[@MrRyanroberson1]

ok wtf youtube unsubscribed me from your channel??

[@PapaFlammy69]

wtf? why does it do that? ;_;

[@MrRyanroberson1]

​@@PapaFlammy69 i don't even know! i noticed was because it was suggesting your vids to me... as a "new channel" to check out for some reason. honestly though i'm afraid for the channels i didn't notice...

[@PapaFlammy69]

well...

[@djcoolvibes]

And if you keep applying digitsum, you will always end up with 9

[@angel-ig]

Yeah, because "infinite" digit sum of n is n mod 9

[@ezras7997]

Can you do tjis?!

[@АлександрКовалев-о7я]

def censored_sum(n): ---> s = 0 while n != 0: --2> s += n % 10 n //= 10

[@АлександрКовалев-о7я]

if I understand your problem it right

[@Observer_detector]

dude is that Putnam problem? intersting

[@PapaFlammy69]

Seems to be an AIME Problem! :)

[@АлександрКовалев-о7я]

Maybe, they not understand, that you solve the genial challenges, as Mathmann, but in our "world" we can solve ones another way?

[@АлександрКовалев-о7я]

10:50

[@maxmusterman3371]

daim i just cant watch your videos without headset because the cringe amount is too high in the first 30 seconds

[@devan2604]

first