If the coefficient of $x^{10}$ in the binomial expansion of $\left(\frac{\sqrt{x}}{5^{\frac{1}{4}}}+\frac{\sqrt{5}}{x^{\frac{1}{3}}}\right)^{60}$ is $5^k l$, where $l, k \in N$ and $l$ is co- prime to...

If $\sum_{k=1}^{31}\left({ }^{31} C_k\right)\left({ }^{31} C_{k-1}\right)-\sum_{k=1}^{30}\left({ }^{30} C_k\right)\left({ }^{30} C_{k-1}\right)=\frac{\alpha(60!)}{(30!)(31!)}$, Where $\alpha \in R$, then the value of $16 \alpha$ is equal to

Let $C_r$ denote the binomial coefficient of $x^r$ in the expansion of $(1+x)^{10}$. If $...

The remainder when $(2021)^{2023}$ is divided by 7 is :

If $\frac{1}{2 \cdot 3^{10}}+\frac{1}{2^2 \cdot 3^9}+\ldots \cdot \frac{1}{2^{10} \cdot 3}=\frac{\mathrm{K}}{2^{10} \cdot 3^{10}}$, then the remainder when K is divided by 6 is

The remainder when $3^{2022}$ is divided by 5 is

$25^{190}-19^{190}-8^{190}+2^{190}$ is divisible by

The remainder, when $7^{103}$ is divided by 17 is $\_\_\_\_$ $-$

The product of the last two digits of $(1919)^{1919}$ is $\_\_\_\_$

The number of integral terms in the expansion of $\left(5^{\frac{1}{2}}+7^{\frac{1}{8}}\right)^{1016}$ is