Notes on the 3b1b video "Counting Problems at the Olympiad Level"

This article is a note on the video [official bilingual] Olympiad-level counting questions released by 3Blue1Brown on station B.

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Question : Set S = { 1 , 2 , 3 , ⋯ , 2000 } S=\{1,2,3,\cdots,2000\}S={ 1,2,3,⋯,2000 } How many subsets have the sum of$Multiples\; of\; 5$ ? Empty sets are also included. For example, the subset{ 2 , 3 , 5 } \{2,3,5\}{ 2,3,The sum of 5 } is$2+3+5=10$ , yes$Multiples\; of\; 5$ .

Solution : Consider the polynomial $$\begin{array}{c}f(x)=(1+x^1)(1+x^2)(1+x^3)\cdots (1+x^{2000})\end{array}$$Expand it, $$\begin{array}{c}f(x)=1+x+x^2+2x{\_}^3+2x{\_}^4+3x{\_}^5+\cdots +x^{2001000}\end{array}$$则$x^{5k}(k\in The\; sum\; of\; the\; coefficients\; of\; the\; N)$ items is the answer we require. The principle is very simple, consider the expansion$\left(2\right)$ is how to get, we need in the product formula( 1 ) (1)Each product term in$($$1$$)$$1$ or$x^m$ ) multiplied, eg$x^3$ can be chosen like this$$(1+x^1)(1+x^2)(1+x^3)\cdots (1+x^{2001000})$$can also be selected like this$$(1+x^1)(1+x^2)(1+x^3)\cdots (1+x^{2001000}),$$there are two selection methods, so the expansion formula$\left(2\right)$ in$x^{The\; coefficient\; of\; 3}$ terms is$2$ . These two selection methods correspond to the setSSTwo subsets of$S$ { 1 , 2 } \{1,2\}{ 1,2 } and$\left\{3\right\}$ , where the elements of the set are$The\; exponent\; of\; x$ , the sum of both sets is$3$ , soSSThe sum of two subsets of$S$$3$。

The problem is that it is impossible for us to hard calculate the expansion, so how to find $x^{5k}(k\in$What about the sum of the coefficients of the$N$$)\; term?$Consider introducing$5th$ time$oh=e^{\frac{2\pi i}{5}}=\cos \frac{2}{5}+i\sin \frac{2}{5}$(The letters used in the original video are $\zeta$ , but in China it is generally customary to use$\omega$），则${oh}^5=1$，${({oh}^2)}^5$、${({oh}^3)}^5$、${({oh}^4)}^5$ is also equal to$1$ . The unit root also has a very good property, let$s$ is an integer, consider the sum${({oh}^0)}^s+{({oh}^1)}^s+{\left({oh}^2\right)}^s+{\left({oh}^0\right)}^s+{\left({oh}^4\right)}^s$ , when$s$ is5 5When multiples of$5$$1$ , the sum is obviously$5$ ; when$s$ is not$5$的倍数时，$$\begin{array}{rl}{({oh}^0)}^s+{({oh}^1)}^s+{({oh}^2)}^s+{\left({oh}^0\right)}^s+{({oh}^4)}^s& ={\left({oh}^0\right)}^s\frac{1-{({oh}^s)}^5}{1-{oh}^s}\\ & =\frac{1-1}{1-oh}\\ & =0\end{array}$$So, ${({oh}^0)}^s+{({oh}^1)}^s+{({oh}^2)}^s+{\left({oh}^0\right)}^s+{({oh}^4)}^s=\{\begin{array}{ll}0,& s\text{is not}\text{a multiple of}5\\ 5,& s\text{is}\end{array}$.
Now we will $f(x)$表示成$$f(x)=c_0+c_1{x}^1+c_2{x}^2+c_3{x}^3+\cdots +c_{2001000}{x}^{2001000}$$ where$c_k$is $x^{}$Coefficient of${}^{k\; term.}$Will${oh}^0,{oh}^1,{oh}^2,{oh}^3,{oh}^4$ Substitute into$f\left(x\right)$ gets$$\begin{array}{rl}f({oh}^0)& =c_0+c_1{\left({oh}^0\right)}^1+c_2{\left({oh}^0\right)}^2+c_3{\left({oh}^0\right)}^3+c_4{\left({oh}^0\right)}^4+c_5{\left({oh}^0\right)}^5+\cdots +c_{2001000}{\left({oh}^0\right)}^{2001000}\\ f({oh}^1)& =c_0+c_1{\left({oh}^1\right)}^1+c_2{\left({oh}^1\right)}^2+c_3{\left({oh}^1\right)}^3+c_4{\left({oh}^1\right)}^4+c_5{\left({oh}^1\right)}^5+\cdots +c_{2001000}{\left({oh}^1\right)}^{2001000}\\ f({oh}^2)& =c_0+c_1{\left({oh}^2\right)}^1+c_2{\left({oh}^2\right)}^2+c_3{\left({oh}^2\right)}^3+c_4{\left({oh}^2\right)}^4+c_5{\left({oh}^2\right)}^5+\cdots +c_{2001000}{\left({oh}^2\right)}^{2001000}\\ f({oh}^3)& =c_0+c_1{\left({oh}^3\right)}^1+c_2{\left({oh}^3\right)}^2+c_3{\left({oh}^3\right)}^3+c_4{\left({oh}^3\right)}^4+c_5{\left({oh}^3\right)}^5+\cdots +c_{2001000}{\left({oh}^3\right)}^{2001000}\\ f({oh}^4)& =c_0+c_1{({oh}^4)}^1+c_2{\left({oh}^4\right)}^2+c_3{\left({oh}^4\right)}^3+c_4{\left({oh}^4\right)}^4+c_5{\left({oh}^4\right)}^5+\cdots +c_{2001000}{\left({oh}^4\right)}^{2001000}\end{array}$$Add them up and notice only $The\; exponent\; of\; x$ is$Only\; items\; that\; are\; multiples\; of\; 5$ can be left, and the rest are eliminated, so we have$$\begin{array}{c}f\left({oh}^0\right)+f\left({oh}^1\right)+f\left({oh}^2\right)+f\left({oh}^3\right)+f\left({oh}^4\right)=5(c_0+c_5+c_{10}+\cdots +c_{2001000})\end{array}$$So what we want is $c_0+c_5+c_{10}+\cdots +c_{2001000}=\frac{f\left({oh}^0\right)+f\left(o^1\right)+f\left(o^2\right)+f\left(o^3\right)+f(o^4)}{5}$。

Obviously, $f({oh}_0)=f(1)=2^{2000}$。 considerf$f({oh}^1)=(1+o)(1+{oh}^2)(1+{oh}^3)(1+{oh}^4)(1+{oh}^5)(1+{oh}^6)\cdots (1+{oh}^{2000})$,then${oh}^{5k+r}={oh}^r$ , so it justrepeats theorange$400$次，即$$f({oh}^1)={\left[(1+o)(1+{oh}^2)(1+{oh}^3)(1+{oh}^4)(1+{oh}^5)\right]}^{The\; 400}$$ question is how to findthe orangepart. Don't forget, the root of unit is called the root of unit because${oh}^0,{oh}^1,{oh}^2,{oh}^3,{oh}^4$ is the equation$x^5-1=0$ in the field of complex numbers$5$ roots, so according to the factor theorem,$$x^5-1=(x-{oh}^0)(x-{oh}^1)(x-{oh}^2)(x-{oh}^3)(x-{oh}^4)$$Put$x=-1$代入得$$-2=(-1-{oh}^0)(-1-{oh}^1)(-1-{oh}^2)(-1-{oh}^3)(-1-{oh}^4)$$and the right end is exactlythe orangepart, sothe value of theorange$2$，$f({oh}^1)=2^{400}$。

vs for ${oh}^2,{oh}^3,{oh}^4$ , we think like this: For${oh}^j(j=2,3,4)$，$({oh}^j{)}^1,({oh}^j{)}^2,({oh}^j{)}^3,({oh}^j{)}^4,({oh}^j{)}^5$ of$The\; 5th$ power is also equal to$1$（即${\left[({oh}^j{)}^k\right]}^5=1$，$k=1,2,3,4,5$），所以$({oh}^j{)}^1,({oh}^j{)}^2,({oh}^j{)}^3,({oh}^j{)}^4,({oh}^j{)}^5$ is also the equation$x^5-1=The\; five\; roots\; of\; 0$ , in fact they are${oh}^0,{oh}^1,{oh}^2,{oh}^3,{oh}^{}$Just another permutation of${}^{4\; .}$Therefore corresponding to$f({oh}^{}$The value of the orangepartof${}^j$$)$$2$ For example,$$\begin{array}{rl}f({oh}^0)& =2^{2000}\\ f({oh}^1)& =f({oh}^2)=f({oh}^3)=2^{400}\end{array}$$故$c_0+c_5+c_{10}+\cdots +c_{2001000}=\frac{2^{2000}+4\times 2^{400}}{5}$, ie $The\; sum\; of\; S$ is5 5The number of subsets of multiples of$5$$\frac{2^{2000}+4\times 2^{400}}{5}$。