Regular math vs competition math

Let's try a regular math problem and a Princeton competition math problem.
Regular math: Compute the sum of all positive divisors of the number 186
Competition math: Compute the sum of all positive integers whose positive divisors sum to 186.
This competition problem is from the 2022 Princeton University Math Competition, the number theory section. As you will see, the competition problem requires way more math knowledge and thinking rather than just a standard routine process. This is also a reason why it is more fun to solve math competition problems.

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AP-IP Ben Delo Marcelo Silva Ehud Ezra 3blue1brown Joseph DeStefano
Mark Mann Philippe Zivan Sussholz AlkanKondo89 Adam Quentin Colley
Gary Tugan Stephen Stofka Alex Dodge Gary Huntress Alison Hansel
Delton Ding Klemens Christopher Ursich buda Vincent Poirier Toma Kolev
Tibees Bob Maxell A.B.C Cristian Navarro Jan Bormans Galios Theorist
Robert Sundling Stuart Wurtman Nick S William O'Corrigan Ron Jensen
Patapom Daniel Kahn Lea Denise James Steven Ridgway Jason Bucata
Mirko Schultz xeioex Jean-Manuel Izaret Jason Clement robert huff
Julian Moik Hiu Fung Lam Ronald Bryant Jan Řehák Robert Toltowicz
Angel Marchev, Jr. Antonio Luiz Brandao SquadriWilliam Laderer Natasha Caron Yevonnael Andrew Angel Marchev Sam Padilla ScienceBro Ryan Bingham
Papa Fassi Hoang Nguyen Arun Iyengar Michael Miller Sandun Panthangi
Skorj Olafsen Riley Faison Rolf Waefler Andrew Jack Ingham P Dwag Jason Kevin Davis Franco Tejero Klasseh Khornate Richard Payne Witek Mozga Brandon Smith Jan Lukas Kiermeyer Ralph Sato Kischel Nair Carsten Milkau Keith Kevelson Christoph Hipp Witness Forest Roberts Abd-alijaleel Laraki Anthony Bruent-Bessette Samuel Gronwold Tyler Bennett christopher careta Troy R Katy Lap C Niltiac, Stealer of Souls Jon Daivd R meh Tom Noa Overloop Jude Khine R3factor. Jasmine Soni L wan na Marcelo Silva Samuel N Anthony Rogers Mark Madsen Robert Da Costa Nathan Kean Timothy Raymond Gregory Henzie Lauren Danielle Nadia Rahman Evangline McDonald Yuval Blatt Zahra Parhoun Hassan Alashoor Kaakaopuupod bbaa Joash Hall Andr3w11235 Cadentato Joe Wisniewski Eric Maximilian Mecke
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Thank you all!

[blackpenredpen]

0:22 I blame the autocorrect for not catching my typo 😆

[Qwerty-qxts]

A (slightly) easier way to find if
186 = 1 + p + p² + ... + pⁿ
is possible, is by noticing that p mist divide 186-1 = 5 * 37, so p is either 5 or 37.

[Sphinxycatto]

I like that whenever I'm lacking and not studying, bprp comes with a new video and I feel like to do math again

[nanamacapagal]

10:16
There is a fast way!
Observe the following:
186 should be written as 1 + p + p^2 + ... + p^a
Notice after that first +1 term every term has a factor of p
So just test the prime factors of 185 (5 and 37) and you're set!
In general, if you're looking to represent a factor F in the form 1 + p + p^2 + ... + p^a, the only values you need to test are the prime factors of F-1

[joshuab]

I always liked math because of little things like this, it turns illegible information into knowledge, you can see how people from the past viewed math with a level of mysticism.

[henridelagardere]

Some are blackpilled, others are redpilled. We are happy to be blackandredpenned.

[rufflesinc]

at a certain level, it feels like the problems are hard not because the problem is innately hard, but just deliberately created hard so fewer people can solve them

[advayamshud]

your way of solving is very interesting and worth watching,
thanks bprp!!!

[the.silent.flautist]

Reference is 10:54 for 6x31: weirdly 31 can also be written as 5^0 + 5^1 + 5^2 ... so once you consider 5, you can't consider it again ... I don't understand the part of logic that I am missing ...

[anveshshekhar]

Last time i was this early he didn't even add a +c

[Fematika]

I proctored for this competition! Didn't really get to think about the problems much though.

[canaDavid]

For the 186 case, notice that after subtracting one the RHS divides the prime, so 185=5*37 must divide it. Neither 5 nor 37 works.

[Ninja]

I was just thinking, must the prime in each bracket be distinct? Cuz for the 6*31 i was thinking that 31 = 1+5+5^2, but since we already used 5 in 6=1+5, idk if its valid.

[-petrichor-]

Not sure if anyone will see this but I have devised a formula for a^3-b^3 which I believe is easier to use and can help solve a lot of different types of problems.

The original formula: (a-b)(a^2+b^2+ab)

My formula: x(x^2+3bx+3b^2) where x = a-b

The reason I believe this formula is better is because it doesn’t rely on the variable a, if they ask a problem to find a while giving a-b and a^3-b^3 this could make the solving process easier because when I tried to solve it then traditional way it took a lot of time.

This also is useful for finding a^3-b^3 by hand.
Let a=2763 and b=2760
3(2763^2+2760^2+2763*2760) vs 3(9+9*2760+2760^2)

In the traditional way, you have to find 2763*2763, 2760*2760 and 2763*2760 which would take forever but in my way you just need to find 2760^2 and 2763*3 which is way easier because as I timed it and it took me 5x longer to do it the traditional way.

If you have any opinion on this formula and have any changes please let me know in the reply section!

[canaDavid]

For the 186 case, notice that after subtracting one the RHS divides the prime, so it must be a divisor of 185=5*37. Neither of these work.

[varunanbalasubramanian]

idk how well it could work but to check if a number can be written as 1+x+x^2...x^n we can subtract 1 from the number and fsctorize it.. to narrow down the possibilities, because if we take 186= 1+x+x^2+x^3...x^n then 185 = x+x^2+x^3..x^n, so 185 is always a multiple of x.. so 185 has only 2 factors of 5 and 37.. and 37^2 is way greater than 185, and 5 doesnt work as he said, idk how much easuer this makes it but maybe(?)

[davidesguevillas]

For 10:30 you could use the formula for a geometric series (1-q^n+1)/(1-q) equal 186 and solve for q to see if there is a prime.

[suspended]

In the beginning why did he stop at 6, ie how do you decide when to stop looking for divisors

[jacemandt]

Is it possible that one of the factors (like 62, in this example) could theoretically be expressed as the sum of powers of primes in that manner, but in two different ways? In other words, maybe finding (1+2)(1+61) isn't quite finishing off the case of 3·62?

[جمالزوانه-خد]

Hello, can I communicate with you? I need to solve a differential equation that has been highly controversial by senior professors