Question

If $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$, then the remainder is ______.

Answer 1

$\Rightarrow$ $2^{100}+3^{100}+4^{100}+5^{100}÷7$

$\Rightarrow \frac{(2^6{)}^{16}\cdot (2^4)}{7}+\frac{(3^6{)}^{16}\cdot 3^4}{7}+\frac{(4^6{)}^{16}\cdot 4^4}{7}+\frac{(5^6{)}^{16}\cdot 5^4}{7}$

[using Former's rule for prime members $\frac{a^{p-1}}{p}=$ Remainder $\left(1\right)$ where p is prime no. and a & p are co-primes]

$\therefore \frac{(2^6{)}^{16}}{7}=Rm\left(\frac{1^{16}}{7}\right)=Rm\left(\frac{1}{7}\right)=Rm\left(1\right)$

$\Rightarrow \frac{2^4+3^4+4^4+5^4}{7}$

$=\frac{16+81+256+625}{7}$

$=\frac{978}{7}$

$=$Remainder $\left(5\right)$.