Question

Find remainder of:
10^1283/514
Pls explain WITH STEPS(DETAILED)

Answer 1

I found the best answer for you.

First, make two parts.

By using multiplication of congruence,

$R(\frac{2^{1283}\times5^{1283}}{2\times257})$ R stands for remainder

$R(\frac{2}{2} \times\frac{2^{1282}}{257} \times\frac{5^{1283}}{257})$

$2\times R(\frac{2^{1282}}{257} \times\frac{5^{1283}}{257})$

→$2\times R(\frac{2^{1282}}{257}) \times R(\frac{5^{1283}}{257})$

Now, use congruence.

Since 2⁸≡257≡-1(mod 257)

→2¹⁶≡1(mod 257)

Exponents :- 1282≡2(mod 16)

2¹²⁸²≡2²(mod 257)

2¹²⁸²≡4(mod 257)

∴$R(\frac{2^{1282}}{257})=8$

According to Fermat's little theorem, $a^{p-1}$ ≡ 1(mod $p$)

→5²⁵⁶≡1(mod 257)

Exponents :- 1283≡3(mod 256)

5¹²⁸³≡5³(mod 257)

5¹²⁸³≡125(mod 257)

∴$R(\frac{5^{1283}}{257})=125$

$2\times R(\frac{2^{1282}}{257}) \times R(\frac{5^{1283}}{257})$ ≡ $2\times 4\times 125$ = 1000 ≡ 486(mod 514)