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Question Number 199864 by cortano12 last updated on 10/Nov/23

Find the remainder Σ_(n=1) ^(2019) n^4 when divide by 53

$$\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{2019}} {\sum}}\mathrm{n}^{\mathrm{4}} \:\mathrm{when} \\ $$$$\:\:\mathrm{divide}\:\mathrm{by}\:\mathrm{53}\: \\ $$

Answered by deleteduser1 last updated on 10/Nov/23

Σ_(n=1) ^(2019) n^4 =((2019(2020)[2(2019)+1][3(2019)^2 +3(2019)−1])/(30)) ≡((5(6)[2(5)+1][3(5)^2 +3(5)−1]=30(11)(89)≡^(53) 8)/(30)) ⇒(8/(30))≡x(mod 53)⇒15x≡4(mod 53)⇒x≡^(53) 25

$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{2019}} {\sum}}{n}^{\mathrm{4}} =\frac{\mathrm{2019}\left(\mathrm{2020}\right)\left[\mathrm{2}\left(\mathrm{2019}\right)+\mathrm{1}\right]\left[\mathrm{3}\left(\mathrm{2019}\right)^{\mathrm{2}} +\mathrm{3}\left(\mathrm{2019}\right)−\mathrm{1}\right]}{\mathrm{30}} \\ $$$$\equiv\frac{\mathrm{5}\left(\mathrm{6}\right)\left[\mathrm{2}\left(\mathrm{5}\right)+\mathrm{1}\right]\left[\mathrm{3}\left(\mathrm{5}\right)^{\mathrm{2}} +\mathrm{3}\left(\mathrm{5}\right)−\mathrm{1}\right]=\mathrm{30}\left(\mathrm{11}\right)\left(\mathrm{89}\right)\overset{\mathrm{53}} {\equiv}\mathrm{8}}{\mathrm{30}} \\ $$$$\Rightarrow\frac{\mathrm{8}}{\mathrm{30}}\equiv{x}\left({mod}\:\mathrm{53}\right)\Rightarrow\mathrm{15}{x}\equiv\mathrm{4}\left({mod}\:\mathrm{53}\right)\Rightarrow{x}\overset{\mathrm{53}} {\equiv}\mathrm{25} \\ $$

Commented by cortano12 last updated on 10/Nov/23

i got 25

$$\mathrm{i}\:\mathrm{got}\:\mathrm{25} \\ $$

Commented by deleteduser1 last updated on 10/Nov/23

≡38×2(1^4 +2^4 +3^4 +4^4 +5^4 +6^4 +...+25^4 +26^4 ) +1^4 +2^4 +3^4 +4^4 +5^4 ≡38×2(25+6^4 +7^4 +...+26^4 ) +25≡38×2(25+27+21+44+42)+25≡^(53) 25 something like this?

$$\equiv\mathrm{38}×\mathrm{2}\left(\mathrm{1}^{\mathrm{4}} +\mathrm{2}^{\mathrm{4}} +\mathrm{3}^{\mathrm{4}} +\mathrm{4}^{\mathrm{4}} +\mathrm{5}^{\mathrm{4}} +\mathrm{6}^{\mathrm{4}} +...+\mathrm{25}^{\mathrm{4}} +\mathrm{26}^{\mathrm{4}} \right) \\ $$$$+\mathrm{1}^{\mathrm{4}} +\mathrm{2}^{\mathrm{4}} +\mathrm{3}^{\mathrm{4}} +\mathrm{4}^{\mathrm{4}} +\mathrm{5}^{\mathrm{4}} \equiv\mathrm{38}×\mathrm{2}\left(\mathrm{25}+\mathrm{6}^{\mathrm{4}} +\mathrm{7}^{\mathrm{4}} +...+\mathrm{26}^{\mathrm{4}} \right) \\ $$$$+\mathrm{25}\equiv\mathrm{38}×\mathrm{2}\left(\mathrm{25}+\mathrm{27}+\mathrm{21}+\mathrm{44}+\mathrm{42}\right)+\mathrm{25}\overset{\mathrm{53}} {\equiv}\mathrm{25} \\ $$$$ \\ $$$${something}\:{like}\:{this}? \\ $$

Commented by cortano12 last updated on 11/Nov/23

yes

$$\mathrm{yes} \\ $$

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