Can you Find the Remainder (2)?

Since, $53$ is a PRIME number, so we can apply Fermat’s “little” Theorem which states that $a^{p-1}\equiv 1\pmod{p}$ where $p$ is a PRIME and $p\nmid a$

Hence, by Fermat’s “little” Theorem we get:

$2^{53-1}=2^{52}\equiv 1\pmod{53}$

Now by raising both sides by $2$ and multiplying both sides by $2^{-2}$ we get: $(2^{53})^2\times 2^{-2}=2^{102}\equiv 1^2\times 2^{-2}=2^{-2}\pmod{53}$

$\implies 2^{102}\equiv (2^{-1})^2\pmod{53}$

$\implies 2^{102}\equiv 27^2\pmod{53}$

$\implies 2^{102}\equiv 729\pmod{53}$

$\implies 2^{102}\equiv 40\pmod{53}$

Hence, the remainder is $\boxed{40}$

Footnotes:-

• How to solve modulo with negative exponents (an example) - youtube .

By Fermat's little theorem , $2^{52} \bmod {53} = 1$ , so $2^{102} \bmod {53} = 2^{102 \bmod {52}} \bmod{53} = 2^{50} \bmod{53} = 2^{-2} \cdot 2^{52} \bmod {53} = 2^{-2} \bmod{53}.$

We are essentially solving for the smallest positive integer $x$ satisfying $4 x \equiv 1 \pmod{53}$ .

I want to make the right hand side of the equivalence, to be divisible by 4, so that I can divide the left hand side by 4:

$\begin{aligned} 4x& \equiv &(1 + 53\times3) \pmod{53} \\ 4x & \equiv & 160 \pmod{53} \\ x & \equiv & 40 \pmod{53} \end{aligned}$

The answer is $\boxed{40}$ .