Download App

STD 11 - 7. binomial theoram — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 7. binomial theoram4 Marks
Question
The greatest integer which divides the number ${101^{100}} - 1$, is
✓

Answer

c
(c) ${(1 + 100)^{100}} = 1 + 100.100 + \frac{{100.99}}{{1.2}}.{(100)^2} + \frac{{100.99.98}}{{1.2.3}}{(100)^3} + ....$

${(101)^{100}} - 1 = 100.100\left[ {1 + \frac{{100.99}}{{1.2}} + \frac{{100.99.98}}{{1.2.3}}.100 + ....} \right]$

From above it is clear that,

${(101)^{100}} - 1$ is divisible by $(100)^2$ $= 10000$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The number of real roots of the equation $\mathrm{e}^{4 \mathrm{x}}+2 \mathrm{e}^{3 \mathrm{x}}-\mathrm{e}^{\mathrm{x}}-6=0$ is :
→
For a polynomial $g ( x )$ with real coefficient, let $m _{ g }$ denote the number of distinct real roots of $g ( x )$. Suppose $S$ is the set of polynomials with real coefficient defined by

$S=\left\{\left(x^2-1\right)^2\left(a_0+a_1 x+a_2 x^2+a_3 x^3\right): a_0, a_1, a_2, a_3 \in R\right\} \text {. }$

For a polynomial $f$, let $f^{\prime}$ and $f^{\prime \prime}$ denote its first and second order derivatives, respectively. Then the minimum possible value of $\left(m_f+m_{f^{\prime}}\right)$, where $f \in S$, is. . . . . . . .

→
Let $P=\{z \in:|z+2-3 i | \leq 1\}$ and $Q=\{z \in C : z(1+i)+\bar{z}(1-i) \leq-8\}$. Let in $P \cap Q ,| z -3+2 i |$ be maximum and minimum at $z_1$ and $z_2$ respectively. If $\left|z_1\right|^2+2|z|^2=\alpha+\beta \sqrt{2}$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ equals $..........$
→
Suppose $\quad f : R \rightarrow(0, \infty)$ be a differentiable function such that $5 f ( x + y )= f ( x ) \cdot f ( y ), \forall x , y \in R$. If $f(3)=320$, then $\sum \limits_{n=0}^5 f(n)$ is equal to :
→
Let $\vec a\, = \,\hat i\, + \,\hat j\, + \,\sqrt 2 \hat k,\,\,\vec b\, = \,{b_1}\hat i\, + \,{b_2}\hat j\, + \sqrt 2 \hat k$ and $\vec c\, = \,5\hat i\, + \,\hat j + \sqrt 2 \hat k$ be three vectors such that the projection vector of $\vec b$ on $\vec a$ is $\vec a$. If $\vec a\, + \vec b$ is perpendicular to $\vec c$ , then $\left| {\vec b} \right|$ is equal to
→
In a class of $100$ students,$15$ students chose only physics (but not mathematics and chemistry),$3$ chose only chemistry (but not mathematics and physics), and $45$ chose only mathematics(but not physics and chemistry). Of the remaining students, it is found that $23$ have taken physics and chemistry,$20$ have taken physics and mathematics, and $12$ have taken mathematics and chemistry. The number of student who chose all the three subjects is
→
The population of cattle in a farm increases so that the difference between the population in year $n+2$ and that in year $n$ is proportional to the population in year $n+1$. If the populations in years $2010, 2011$ and $2013$ were $39,60$ and $123$,respectively, then the population in $2012$ was
→
The value of $\int_0^1\left(2 x^3-3 x^2-x+1\right)^{\frac{1}{3}} d x$ is equal to:
→
The sides $AB, BC, CA$ of a triangle $ABC$ have respectively $3, 4$ and $5$ points lying on them. The number of triangles that can be constructed using these points as vertices is
→
If $u = {\log _e}({x^2} + {y^2}) + {\tan ^{ - 1}}\left( {{y \over x}} \right)$, then ${{{\partial ^2}u} \over {\partial {x^2}}} + {{{\partial ^2}u} \over {\partial {y^2}}} = $
→