UPSC CDS 1 2022 Math Paper

Given : $p=2^{2n+2}+m$ and $q=2^{4n}-m$ $n$ is even natural number Formula used : $(M^{ab})=(M^{a})^{b}$ $(M^{a+b})=(M^{a} \cdot M^{b})$ Calculations : If $p$ is divisible by 5 , we have $\Rightarrow \frac{p}{5}=\frac{2^{2n+2}+m}{5}$ Using the formula above $\Rightarrow \frac{(2^{2})^{n} \times 2^{2}+m}{5}$ $\Rightarrow \frac{4^{n} \times 4+m}{5}$ When 4 is divided by 5 it leaves - 1 as a negative remainder, so $\Rightarrow \frac{(-1^{n}) \times 4+m}{5}$ $n$ is an even positive number $\Rightarrow \frac{1 \times 4+m}{5}$ $\Rightarrow \frac{4+m}{5}$ The least number required for $m$ is 1 $\Rightarrow \frac{4+1}{5}=\frac{5}{5}=1$ Also, we have $q=2^{4n}-m$ $\Rightarrow q=16^{n}-m$ Since $n$ is an even positive integer. Therefore, (16-m) will always be one factor of q. If $m=1$ then $q=16-1=15$ Which is divisible by 5 . $\therefore$ The least value of $m$ such that $p$, as well as $q$, is divisible by 5 is 1 .