Great point. The description already expands on the conditions for the statement that I couldn't fit in the thumbnail. There's a non trivial expansion for each N where A and C are square free. You could disallow some trivial expansions like you presented and the question is whether any other non trivial expansions exist to which the video answers in affirmative
@@numerodivergence you're correct but there's always the problem with prime numbers which have to be expressed as A*1²*1³. Since you can restrict that, you end up being able to express all numbers this way
1. 5 threes can become 3 fives for free 2. 3 threes + 1 => 2 fives 3. 1 five + 1 => 2 threes therefore a continuum is reached at and after 8 (1 three + 1 five => 3 threes => induction) where the 3 threes => 2 fives => 4 threes => 6 fives => ... 2 2^n threes => 3 2^n fives => ... a special situation occurs for 7 and below where there aren't enough fives or threes to increment without using negative coins (7 = 10-3, 4 = 10-6, 2 = 5-3), which is usually possible if the person you're making change to has the coins needed to make the negative value.
Thank you for the comment. Yeah, that's right! I didn't talk about generating functions which you mention. You also get a nice formula for the generating function as a reciprocal of a quadratic polynomial
6:50 even simpler, all the even exponents go to the f^2 and of the odd exponents 3 go with g&3 and the rest go with f^2. So if n = a^2 * b^3 * c^4 * d^5, p would be a * c^2 * d, and g would be b * d.
That's true, and your process is basically the algo in the video showcased for the simple case. The video tackles the general question and the question posed is a very simple corollary as you pointed out
Ohh, "power-ful". Full of powers. Very good. *Sigh and facepalm* why did I expect something better from mathematicians (disclaimer: I'm a mathematician)
Your result that 'there exists z such that A z = 1 (mod b)' is only true for all A if b is prime. For example, A = 2 b = 4 has no such solution for z. So I think the proof must be more complex than what you've shown.
Hi! Thanks for the comment. The hypothesis is that "a and b are coprime" where coprime means that they share no factors other than 1 ofc. 2 and 4 share the factor 2 and aren't coprime So when they don't share any factors like 14 and 15, the result is true Your statement is actually stronger in the sense that *all* numbers mod b have inverses when b is prime. A more general form of that result is the coprime one stated in the video! Thanks for bringing this up
1:13 "Throughout this video, we will assume A and B are coprime" His result that 'there is z such that Az=1 (mod B)' is TRUE because he said gcd(A,B)=1. "Throughout this video ... A and B are coprime"
i agree, but mumble rap doesn't exist: the music you're talking about is actually a plethora of sub-genres like trap. the term "mumble rap" was created by white people who do not understand rap as a genre or how art works.
"You didn't prove the statement in the thumbnail. Is this clickbait?!!?"
No... it's homework.
(there's a solution in the description)
If A,B,C can be 1, then the problem is just N=N*1²*1³, so I'm a bit confused as to how this is even a question
Great point. The description already expands on the conditions for the statement that I couldn't fit in the thumbnail. There's a non trivial expansion for each N where A and C are square free. You could disallow some trivial expansions like you presented and the question is whether any other non trivial expansions exist to which the video answers in affirmative
@@numerodivergence you're correct but there's always the problem with prime numbers which have to be expressed as A*1²*1³. Since you can restrict that, you end up being able to express all numbers this way
TIL the powerful in powerful numbers doesn't mean they're stronger than other numbers
my number can beat up your number
Frobenius coin problem? Ahh you mean the chicken mcnugget theorem
This probably has many names and no doubt Chicken McNugget is one of the best ones
43 mcnuggets😔@@numerodivergence
Although in the McNugget problem, you can have the gcd of the coin denominations not equal to 1
@@numerodivergence ik but it's funny
Oh my Euler, yes, the delicious reasoning using inverses in mod N is so refreshing, so satisfactory ...
Ah indeed, so good...
Thank You
Thanks for watching!
1. 5 threes can become 3 fives for free
2. 3 threes + 1 => 2 fives
3. 1 five + 1 => 2 threes
therefore a continuum is reached at and after 8 (1 three + 1 five => 3 threes => induction) where the 3 threes => 2 fives => 4 threes => 6 fives => ... 2 2^n threes => 3 2^n fives => ...
a special situation occurs for 7 and below where there aren't enough fives or threes to increment without using negative coins (7 = 10-3, 4 = 10-6, 2 = 5-3), which is usually possible if the person you're making change to has the coins needed to make the negative value.
numbers that can be expressed using 3 & 5 are coefficients > 0 of x^n
in (1+x^3+x^6+...)(1+x^5+x^10+...) ( Combinatorics )
Thank you for the comment. Yeah, that's right! I didn't talk about generating functions which you mention. You also get a nice formula for the generating function as a reciprocal of a quadratic polynomial
@@numerodivergence Yes :)
I'm used to seeing f and g as functions - x/y or a/b as other things.
Loved this ❤
Great video!
Clicked purely cuz of the Hades reference
Hope you weren't disappointed!
6:50 even simpler, all the even exponents go to the f^2 and of the odd exponents 3 go with g&3 and the rest go with f^2. So if n = a^2 * b^3 * c^4 * d^5, p would be a * c^2 * d, and g would be b * d.
That's true, and your process is basically the algo in the video showcased for the simple case. The video tackles the general question and the question posed is a very simple corollary as you pointed out
Ohh, "power-ful". Full of powers. Very good. *Sigh and facepalm* why did I expect something better from mathematicians (disclaimer: I'm a mathematician)
Your result that 'there exists z such that A z = 1 (mod b)' is only true for all A if b is prime. For example, A = 2 b = 4 has no such solution for z. So I think the proof must be more complex than what you've shown.
Hi! Thanks for the comment. The hypothesis is that "a and b are coprime" where coprime means that they share no factors other than 1 ofc. 2 and 4 share the factor 2 and aren't coprime
So when they don't share any factors like 14 and 15, the result is true
Your statement is actually stronger in the sense that *all* numbers mod b have inverses when b is prime. A more general form of that result is the coprime one stated in the video! Thanks for bringing this up
1:13 "Throughout this video, we will assume A and B are coprime"
His result that 'there is z such that Az=1 (mod B)' is TRUE because he said gcd(A,B)=1.
"Throughout this video ... A and B are coprime"
@@samueldeandrade8535 yes I missed this constraint
wait what
I didnt know you could make all numbers given 2 presumably coprime coins
All numbers except a finite few in the beginning
Not saying this video has it but mumble rap counts as music :)
i agree, but mumble rap doesn't exist: the music you're talking about is actually a plethora of sub-genres like trap. the term "mumble rap" was created by white people who do not understand rap as a genre or how art works.
Are we doing random opinions nowadays?
Sorry to whoever took it seriously. This is just an inside joke :)