Xavier Aptitude Test 2020 Paper

Using Fermat's theorem :
If p is a prime number and a, p are co primes ${\mathit{a}}^{\mathit{p}-1}\mathrm{mod}\mathit{p}=1$
Remainder when $\frac{{19}^{20}}{7}={19}^2\mathrm{mod}7=4$
Since the remainder for ${19}^6$ is $1$ the remainder for ${19}^{20}$ is equivalent to the $\frac{{19}^2}{7}=4$
Remainder when $\frac{{20}^{19}}{7}={20}^1\mathrm{mod}7=6$
${20}^{19}={\left({20}^6\right)}^3.{20}^1=\frac{1.20}{7}$
Remainder when ${19}^{20}-{20}^{19}$ is divided by $7=4-6=-2\Rightarrow 5$