Binomial Theorem for Positive Integral Index

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

The number of integral terms in the expansion of ${\left(3^{\frac{1}{2}}+5^{\frac{1}{4}}\right)}^{680}$ is equal to

If the ${1011}^{\mathrm{th}}$ term from the end in the binomial expansion of ${\left(\frac{4x}{5}-\frac{5}{2x}\right)}^{2022}$ is $1024$ times ${1011}^{\mathrm{th}}$ term from the beginning, then $32\left|x\right|$ is equal to

The sum, of the coefficients of the first $50$ terms in the binomial expansion of ${\left(1-x\right)}^{100},$ is equal to

The coefficient of $x^5$ in the expansion of ${\left(2x^3-\frac{1}{3x^2}\right)}^5$ is 

If the coefficients of $x$ and $x^2$ in $(1+x{)}^p(1-x{)}^q$ are $4$ and $-5$ respectively, then $2p+3q$ is equal to

Let $\alpha$ be the constant term in the binomial expansion of ${\left(\sqrt{x}-\frac{6}{x^{\frac{3}{2}}}\right)}^n,n\le 15.$ If the sum of the coefficients of the remaining terms in the expansion is $649$ and the coefficient of $x^{-n}$ is $\lambda \alpha ,$ then $\lambda$ is equal to $\_\_\_\_\_\_\_\_.$

Fractional part of the number $\frac{4^{2022}}{15}$ is equal to

If $\frac{1}{n+1}{ }^n{C}_n+\frac{1}{n}{ }^n{C}_{n-1}+...+\frac{1}{2}{ }^n{C}_1{+}^n{C}_0=\frac{1023}{10}$ then $n$ is equal to

Let the number $(22{)}^{2022 }+ (2022{)}^{22}$ leave the remainder $\mathrm{\alpha }$ when divided by $3$ and $\mathrm{\beta }$ when divided by $7$. Then $({\mathrm{\alpha }}^2 + {\mathrm{\beta }}^2 )$ is equal to

Let $\left[t\right]$ denote the greatest integer $\le t$. if the constant term in the expansion of ${\left(3x^2-\frac{1}{2x^5}\right)}^7$ is $\alpha$ then $\left[\alpha \right]$ is equal to $\_\_\_\_\_$

If the coefficient of $x^7$ in ${\left(ax-\frac{1}{b{x}^2}\right)}^{13}$ and the coefficient of $x^{-5}$ in ${\left(ax+\frac{1}{b{x}^2}\right)}^{13}$ are equal, then $a^4{b}^4$ is equal to: 

Among the statements :

$\left(S1\right) : {2023}^{2022}-{1999}^{2022}$ is divisible by $8$.

$\left(S2\right) : 13(13{)}^n-11n-13$ is divisible by $144$ for infinitely many $n\in \mathrm{\mathbb{N}}$

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

The absolute difference of the coefficients of $x^{10}$ and $x^7$ in the expansion of ${\left(2x^2+\frac{1}{2x}\right)}^{11}$ is equal to

If the coefficients of $x^7$ in ${\left(a{x}^2+\frac{1}{2bx}\right)}^{11}$ and $x^{-7}$ in ${\left(ax-\frac{1}{3b{x}^2}\right)}^{11}$ are equal, then

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)}^n$ is $\sqrt{6}:1$, then the third term from the beginning is:

Find the remainder when ${23}^{200}+{19}^{200}$ is divided by $49$.

Remainder when $2^{2022}$ is divided by $15$ is equal to 

For the expression ${\left(1-x\right)}^{100}$. Then sum of coefficients of first $50$ terms is: