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

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

Answered by deleteduser1 last updated on 10/Nov/23

$$\underset{n=1}{\sum ^{2019}}{n}^4=\frac{2019\left(2020\right)[2\left(2019\right)+1][3{\left(2019\right)}^2+3\left(2019\right)-1]}{30}$$$$\equiv \frac{5\left(6\right)[2\left(5\right)+1][3{\left(5\right)}^2+3\left(5\right)-1]=30\left(11\right)\left(89\right)\stackrel{53}{\equiv }8}{30}$$$$\Rightarrow \frac{8}{30}\equiv x\left(mod53\right)\Rightarrow 15x\equiv 4\left(mod53\right)\Rightarrow x\stackrel{53}{\equiv }25$$

Commented by cortano12 last updated on 10/Nov/23

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

Commented by deleteduser1 last updated on 10/Nov/23

$$\equiv 38\times 2(1^4+2^4+3^4+4^4+5^4+6^4+...+{25}^4+{26}^4)$$$$+1^4+2^4+3^4+4^4+5^4\equiv 38\times 2(25+6^4+7^4+...+{26}^4)$$$$+25\equiv 38\times 2(25+27+21+44+42)+25\stackrel{53}{\equiv }25$$$$$$$$somethinglikethis?$$

Commented by cortano12 last updated on 11/Nov/23

$$\mathrm{yes}$$

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