Question

The number ${{\left( 101 \right)}^{100}}-1$ is divisible by
(A) ${{10}^{4}}$
(B) ${{10}^{6}}$
(C) ${{10}^{8}}$
(D) ${{10}^{12}}$

Answer 1

In the given expression, first of all, replace 101 by $1+100=101$ . Now, use the binomial expansion formula, ${{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+........{{+}^{n}}{{C}_{n}}{{x}^{n}}$ and expand the expression
${{\left( 1+100 \right)}^{100}}-1$ . Use, $^{100}{{C}_{0}}=1$ and $^{100}{{C}_{1}}=100$ . Solve it further and get the number by which the given expression is divisible.

Complete step-by-step solution
According to the question, we are given an expression and we have to find the number by which the given expression is completely divisible.
The given expression = ${{\left( 101 \right)}^{100}}-1$ ………………………………..(1)
Since the above expression is too complex to figure out the number by which it is completely divisible. So, we have to simplify the above expression into a simpler form.
We can write 101 as the summation of 1 and 100, which is $1+100=101$.
Now, using $101+1=101$ and on transforming the expression in equation (1), we get
${{\left( 1+100 \right)}^{100}}-1$ ……………………………….(2)
We know the formula for the binomial expansion, ${{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+........{{+}^{n}}{{C}_{n}}{{x}^{n}}$ …………………………………………………(3)
Now, on putting $x=100$ and $n=100$ in equation (3), we get
$\Rightarrow {{\left( 1+100 \right)}^{100}}{{=}^{100}}{{C}_{0}}{{+}^{100}}{{C}_{1}}\left( 100 \right){{+}^{100}}{{C}_{2}}{{\left( 100 \right)}^{2}}+........{{+}^{100}}{{C}_{100}}{{\left( 100 \right)}^{100}}$ …………………………………………….(4)
Using equation (4) and on substituting ${{\left( 1+100 \right)}^{100}}$ by $^{100}{{C}_{0}}{{+}^{100}}{{C}_{1}}100{{+}^{100}}{{C}_{2}}{{\left( 100 \right)}^{2}}+........{{+}^{100}}{{C}_{100}}{{\left( 100 \right)}^{100}}$ in equation (2), we get
${{=}^{100}}{{C}_{0}}{{+}^{100}}{{C}_{1}}100{{+}^{100}}{{C}_{2}}{{\left( 100 \right)}^{2}}+........{{+}^{100}}{{C}_{100}}{{\left( 100 \right)}^{100}}-1$ ……………………………………….(5)
We know that $^{100}{{C}_{0}}=1$ and $^{100}{{C}_{1}}=100$ ……………………………………….(6)
Now, using equation (6) and on replacing $^{100}{{C}_{0}}$ and $^{100}{{C}_{1}}$ by 1 and 100 in equation (5), we get
$=1+100\times 100{{+}^{100}}{{C}_{2}}{{\left( 100 \right)}^{2}}+........{{+}^{100}}{{C}_{100}}{{\left( 100 \right)}^{100}}-1$
On replacing 100 by ${{10}^{2}}$ in the above equation, we get
$=1+{{10}^{2}}\times {{10}^{2}}{{+}^{100}}{{C}_{2}}{{\left( {{10}^{2}} \right)}^{2}}+........{{+}^{100}}{{C}_{100}}{{\left( {{10}^{2}} \right)}^{100}}-1$
$=1+{{10}^{4}}{{+}^{100}}{{C}_{2}}{{\left( 10 \right)}^{4}}+........{{+}^{100}}{{C}_{100}}{{\left( 10 \right)}^{200}}-1$ …………………………………………(7)
In the above equation, we can see that we have 1 and -1, which will cancel each other.
On canceling 1 by -1 in equation (7), we get
$={{10}^{4}}{{+}^{100}}{{C}_{2}}{{\left( 10 \right)}^{4}}+........{{+}^{100}}{{C}_{100}}{{\left( 10 \right)}^{200}}$ ……………………………………….(8)
Now, in equation (8), on taking the term ${{10}^{4}}$ as common, we get
={{10}^{4}}\left\\{1+ ^{100}{{C}_{2}}+........{{+}^{100}}{{C}_{100}}{{\left( 10 \right)}^{200-4}} \right\\}
={{10}^{4}}\left\\{1+ ^{100}{{C}_{2}}+........{{+}^{100}}{{C}_{100}}{{\left( 10 \right)}^{196}} \right\\} …………………………………………(9)
We can see that the above equation is divisible by ${{10}^{4}}$ .
Therefore, the given expression ${{\left( 101 \right)}^{100}}-1$ is divisible by ${{10}^{4}}$.
Hence, the correct option is (A).

Note: Whenever this type of question appears where we are given an expression and we have to find the number by which the given expression is completely divisible. Always approach this type of question by using the binomial expansion formula, ${{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+........{{+}^{n}}{{C}_{n}}{{x}^{n}}$.