Find-the-remainder-n-1-2019-n-4-when-divide-by-53-

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

Find the remainder \underset{n = 1}{\sum^{2019}} n^{4} when

divide by 53

Answered by AST last updated on 10/Nov/23

\underset{n = 1}{\sum^{2019}} n^{4} = \frac{2019 (2020) [2 (2019) + 1] [3 {(2019)}^{2} + 3 (2019) - 1]}{30}

\equiv \frac{5 (6) [2 (5) + 1] [3 {(5)}^{2} + 3 (5) - 1] = 30 (11) (89) \overset{53}{\equiv} 8}{30}

\Rightarrow \frac{8}{30} \equiv x (m o d 53) \Rightarrow 15 x \equiv 4 (m o d 53) \Rightarrow x \overset{53}{\equiv} 25

Commented by cortano12 last updated on 10/Nov/23

i got 25

Commented by AST last updated on 10/Nov/23

\equiv 38 \times 2 (1^{4} + 2^{4} + 3^{4} + 4^{4} + 5^{4} + 6^{4} + \dots + 25^{4} + 26^{4})

+ 1^{4} + 2^{4} + 3^{4} + 4^{4} + 5^{4} \equiv 38 \times 2 (25 + 6^{4} + 7^{4} + \dots + 26^{4})

+ 25 \equiv 38 \times 2 (25 + 27 + 21 + 44 + 42) + 25 \overset{53}{\equiv} 25

s o m e t h i n g l i k e t h i s ?

Commented by cortano12 last updated on 11/Nov/23

yes