JEE 2023 — Mathematics PYQ

Solution

1. Express in terms of 21: Note that $19=21-2$ and $23=21+2$.

   $$(21-2{)}^{200}+(21+2{)}^{200}$$

2. Expand using Binomial Theorem:

   $(21-2{)}^{200}=(\genfrac{}{}{0ex}{}{200}{0})21^{200}-\cdots -(\genfrac{}{}{0ex}{}{200}{199})21(2{)}^{199}+\left(\genfrac{}{}{0ex}{}{200}{200}\right)2^{200}$

   $(21+2{)}^{200}=(\genfrac{}{}{0ex}{}{200}{0})21^{200}+\cdots +(\genfrac{}{}{0ex}{}{200}{199})21(2{)}^{199}+\left(\genfrac{}{}{0ex}{}{200}{200}\right)2^{200}$

3. Summing the terms:

   Sum $=2[(\genfrac{}{}{0ex}{}{200}{0})21^{200}+(\genfrac{}{}{0ex}{}{200}{2})21^{198}{2}^2+\cdots +(\genfrac{}{}{0ex}{}{200}{198})21^2{2}^{198}+2^{200}]$

   Every term containing $21^2$ is divisible by $21^2=441$, which is divisible by $49$.

   Remainder is determined by $2\cdot 2^{200}(\mathrm{mod}49)$.

4. Solve $2^{201}(\mathrm{mod}49)$:

   $2^7=128\equiv 30(\mathrm{mod}49)$

   This part requires further simplification or using $21=3\times 7$. Let's re-examine $19\equiv -2(\mathrm{mod}7)$ and $23\equiv 2(\mathrm{mod}7)$. The sum is $0(\mathrm{mod}7)$.

   Using $19^{200}+23^{200}\equiv (21-2{)}^{200}+(21+2{)}^{200}(\mathrm{mod}49)$:

   Sum $\equiv 2^{200}+2^{200}=2^{201}(\mathrm{mod}49)$

   Using Euler's totient function $\varphi (49)=49(1-1/7)=42$.

   $201=4\times 42+33$.

   $2^{33}=(2^{10}{)}^3\cdot 2^3=(1024{)}^3\cdot 8\equiv (44{)}^3\cdot 8\equiv (-5{)}^3\cdot 8=-1000\equiv 29(\mathrm{mod}49)$.

Answer: 29