Question Number 218997 by hardmath last updated on 17/Apr/25
Answered by MrGaster last updated on 19/Apr/25
![4!Σ_(α+β+γ+δ=4) (([x]^α [x/2]^β [x/4]^γ [x/8]^δ )/(α!β!γ!δ!))=([x]+[(x/2)]+[(x/4)]+[(x/8)])^4 ⇒([x]+[(x/2)]+[(x/4)][(x/8)])^4 ≡1 (mod 5) (^∗ ) if S+[x]+[(x/2)]+[(x/4)]+[(x/8)].Then it is formed by (∗) ofm Ferats theorem. Therefore as long as x isu discssed in categories. Let S=[x]+[(x/2)]+[(x/4)]+[(x/6)] When considering [1,2] ⇓ ∧^([1,2]) S=1(✓) determinant ((([2,3)),(3✓)),(([3,4)),(4✓)),(([4,5)),(7✓)),(([5,6)),(8✓)),(([6,7)),(10)),(([7,8)),(11 ✓)),(([8,9)),(15 )),(([9,10]),(16✓)))& determinant ((([10,11)),(18✓)),(([11,12)),(19✓)),(([12,13)),(22✓)),(([13,14)),(23✓)),(([14,15)),(25 )),(([15,16]),(26)),(([16,17]),(30)))& determinant ((([17,18)),(31✓)),(([18,19)),(33✓)),(([19,20)),(37✓)),(([20,21)),(37✓)),(([21,22)),(38✓)),(([22,23)),(40)),(([23,24)),(41)),(([24]),(45))) ∴ determinant (((x6[1,6)∪[7.8)∪[9,14)∪[15,16)∪[17,22)∪[23,24))))](https://www.tinkutara.com/question/Q219096.png)
$$\mathrm{4}!\underset{\alpha+\beta+\gamma+\delta=\mathrm{4}} {\sum}\frac{\left[{x}\right]^{\alpha} \left[{x}/\mathrm{2}\right]^{\beta} \left[{x}/\mathrm{4}\right]^{\gamma} \left[{x}/\mathrm{8}\right]^{\delta} }{\alpha!\beta!\gamma!\delta!}=\left(\left[{x}\right]+\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{{x}}{\mathrm{4}}\right]+\left[\frac{{x}}{\mathrm{8}}\right]\right)^{\mathrm{4}} \\ $$$$\Rightarrow\left(\left[{x}\right]+\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{{x}}{\mathrm{4}}\right]\left[\frac{{x}}{\mathrm{8}}\right]\right)^{\mathrm{4}} \equiv\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{5}\right)\:\left(^{\ast} \right) \\ $$$$\mathrm{if}\:{S}+\left[{x}\right]+\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{{x}}{\mathrm{4}}\right]+\left[\frac{{x}}{\mathrm{8}}\right].\mathrm{Then}\:\mathrm{it}\:\mathrm{is}\:\mathrm{formed}\:\mathrm{by}\:\left(\ast\right)\:\mathrm{ofm} \\ $$$$\mathrm{Ferats}\:\mathrm{theorem}. \\ $$$$\mathrm{Therefore}\:\mathrm{as}\:\mathrm{long}\:\mathrm{as}\:{x}\:\mathrm{isu} \\ $$$$\mathrm{discssed}\:\mathrm{in}\:\mathrm{categories}. \\ $$$$\mathrm{Let}\:{S}=\left[{x}\right]+\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{{x}}{\mathrm{4}}\right]+\left[\frac{{x}}{\mathrm{6}}\right] \\ $$$$\mathrm{When}\:\mathrm{considering}\:\left[\mathrm{1},\mathrm{2}\right] \\ $$$$\Downarrow\:\:\:\:\:\:\:\:\:\:\:\:\overset{\left[\mathrm{1},\mathrm{2}\right]} {\wedge}{S}=\mathrm{1}\left(\checkmark\right) \\ $$$$\begin{array}{|c|c|c|c|c|c|c|c|}{\left[\mathrm{2},\mathrm{3}\right)}&\hline{\mathrm{3}\checkmark}\\{\left[\mathrm{3},\mathrm{4}\right)}&\hline{\mathrm{4}\checkmark}\\{\left[\mathrm{4},\mathrm{5}\right)}&\hline{\mathrm{7}\checkmark}\\{\left[\mathrm{5},\mathrm{6}\right)}&\hline{\mathrm{8}\checkmark}\\{\left[\mathrm{6},\mathrm{7}\right)}&\hline{\cancel{\mathrm{10}}}\\{\left[\mathrm{7},\mathrm{8}\right)}&\hline{\mathrm{11}\:\checkmark}\\{\left[\mathrm{8},\mathrm{9}\right)}&\hline{\cancel{\mathrm{15}\:}}\\{\left[\mathrm{9},\mathrm{10}\right]}&\hline{\mathrm{16}\checkmark}\\\hline\end{array}\&\begin{array}{|c|c|c|c|c|c|c|}{\left[\mathrm{10},\mathrm{11}\right)}&\hline{\mathrm{18}\checkmark}\\{\left[\mathrm{11},\mathrm{12}\right)}&\hline{\mathrm{19}\checkmark}\\{\left[\mathrm{12},\mathrm{13}\right)}&\hline{\mathrm{22}\checkmark}\\{\left[\mathrm{13},\mathrm{14}\right)}&\hline{\mathrm{23}\checkmark}\\{\left[\mathrm{14},\mathrm{15}\right)}&\hline{\cancel{\mathrm{25}\:}}\\{\left[\mathrm{15},\mathrm{16}\right]}&\hline{\cancel{\mathrm{26}}}\\{\left[\mathrm{16},\mathrm{17}\right]}&\hline{\cancel{\mathrm{30}}}\\\hline\end{array}\&\begin{array}{|c|c|c|c|c|c|c|c|}{\left[\mathrm{17},\mathrm{18}\right)}&\hline{\mathrm{31}\checkmark}\\{\left[\mathrm{18},\mathrm{19}\right)}&\hline{\mathrm{33}\checkmark}\\{\left[\mathrm{19},\mathrm{20}\right)}&\hline{\mathrm{37}\checkmark}\\{\left[\mathrm{20},\mathrm{21}\right)}&\hline{\mathrm{37}\checkmark}\\{\left[\mathrm{21},\mathrm{22}\right)}&\hline{\mathrm{38}\checkmark}\\{\left[\mathrm{22},\mathrm{23}\right)}&\hline{\cancel{\mathrm{40}}}\\{\left[\mathrm{23},\mathrm{24}\right)}&\hline{\mathrm{41}}\\{\left[\mathrm{24}\right]}&\hline{\cancel{\mathrm{45}}}\\\hline\end{array} \\ $$$$\therefore\begin{array}{|c|}{{x}\mathrm{6}\left[\mathrm{1},\mathrm{6}\right)\cup\left[\mathrm{7}.\mathrm{8}\right)\cup\left[\mathrm{9},\mathrm{14}\right)\cup\left[\mathrm{15},\mathrm{16}\right)\cup\left[\mathrm{17},\mathrm{22}\right)\cup\left[\mathrm{23},\mathrm{24}\right)}\\\hline\end{array} \\ $$
Commented by hardmath last updated on 20/Apr/25
$$\mathrm{thankyou}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$