Let $p$ be a prime number, and $a$ be any integer. Then $a^p-a$ is always divisible by $p.$ In modular arithmetic notation, this can be written as $a^p\equiv a \mod p$
Special case: If $a$ is not divisible by $p,$ Fermat's little theorem is equivalent to the statement that $a^{p-1}-1$ is an integer multiple of $p,$ or in symbols $:a^{p-1}\equiv 1(\mod p).$
In other words: Remainder of $\left[\dfrac{a^{p-1}}{p}\right] = 1,$ where $p$ is a prime number and $HCF(a, p) = 1.$
Remainder of$:\dfrac{2^{1040}}{131}$
Here $a = 2,p = 131$ and $HCF(2,131) = 1$
We can write $2^{130} \equiv 1 \mod 131$
$$(OR)$$
$\text{Euler’s Theorem:}$ Consider the number in the form of $\dfrac{a^{p}}{n}$
As per the Euler’s theorem, If $n$ is relatively prime to $a$ then $a^\Phi(n)$ divided by $n$ gives $1$ as the remainder.
i.e Remainder $\left[\dfrac{a^{\Phi(n)}}{n}\right] = 1$
Here, Remainder of$:\dfrac{2^{1040}}{131}$
Here $a = 2,p = 131$ and $HCF(2,131) = 1$
$\Phi(131) = 130\:\left[\because\text{If p is prime, then} \:\Phi(p) = p-1\:\:\text{(or)}\:\: \Phi(p^{n}) = p^{n} – p^{n-1}\right]$
$\therefore \text{Remainder}\:\left[\dfrac{2^{\Phi(131)}}{131}\right] = \left[\dfrac{2^{130}}{131}\right] = 1$
So, the correct answer is $(A).$
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