Question

The last four digits of natural number \[{{7}^{100}}\] are
(a) 2732
(b) 1301
(c) 2500
(d) 0001

Answer 1

Hint: We solve this problem by using the binomial theorem.
We convert the given number in such a way that we get it in the form of \[{{\left( x+1 \right)}^{n}}\] or \[{{\left( x-1 \right)}^{n}}\] where \[x\] is a multiple of 10.
Then we use the binomial expansion of suitable form. The binomial expansion is given as
\[{{\left( x+1 \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}+{}^{n}{{C}_{3}}{{x}^{n-3}}+..........+{}^{n}{{C}_{n}}\]
\[{{\left( x-1 \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}-{}^{n}{{C}_{1}}{{x}^{n-1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}-{}^{n}{{C}_{3}}{{x}^{n-3}}+..........+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}\]
Then we take the suitable terms where we get the last four digits by eliminating the terms that have last four digits as zeros. Then we evaluate the suitable terms to get the last four digits.

Complete step by step answer:
We are asked to find the last four digits of \[{{7}^{100}}\]
Now, let us convert the given number in the form \[{{\left( x+1 \right)}^{n}}\] or \[{{\left( x-1 \right)}^{n}}\] where \[x\] is a multiple of 10.
Here, we can see that the number 7 cannot be written in the required form.
We know that the formula of exponents that is
\[\Rightarrow {{\left( a \right)}^{c\times b}}={{\left( {{a}^{c}} \right)}^{b}}\]

By using the above formula to given number we ge
\[\begin{align}
  & \Rightarrow {{7}^{100}}={{7}^{2\times 50}} \\
 & \Rightarrow {{7}^{100}}={{\left( {{7}^{2}} \right)}^{50}}={{49}^{50}} \\
\end{align}\]
Here, we can see that the number 49 can be written in the required form that is
\[\Rightarrow {{7}^{100}}={{\left( 50-1 \right)}^{50}}\]
We know that the binomial expansion is given as
\[{{\left( x-1 \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}-{}^{n}{{C}_{1}}{{x}^{n-1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}-{}^{n}{{C}_{3}}{{x}^{n-3}}+..........+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}\]
By using this expansion in above equation we get
\[\Rightarrow {{7}^{100}}={}^{50}{{C}_{0}}{{\left( 50 \right)}^{50}}-{}^{50}{{C}_{1}}{{\left( 50 \right)}^{49}}+......-{}^{50}{{C}_{47}}{{\left( 50 \right)}^{3}}+{}^{50}{{C}_{48}}{{\left( 50 \right)}^{2}}-{}^{50}{{C}_{49}}\left( 50 \right)+{}^{50}{{C}_{50}}\]
Here we can see that the power of 50 is reduced from 50 to 0
Now, we can see that the last four digits of the terms up to \[{{50}^{4}}\] will be zeros.
So, we can represent the terms that are up to \[{{50}^{4}}\] as \[10000\times k\] where \[k\] is some integer.
Now, let us rewrite the above expansion by representing the up to \[{{50}^{4}}\] as shown then we get
\[\Rightarrow {{7}^{100}}=\left( 10000\times k \right)-{}^{50}{{C}_{47}}{{\left( 50 \right)}^{3}}+{}^{50}{{C}_{48}}{{\left( 50 \right)}^{2}}-{}^{50}{{C}_{49}}\left( 50 \right)+{}^{50}{{C}_{50}}\]
We know that the formula of combinations that is
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using this formula in above equation we get
     \[\begin{align}
  & \Rightarrow {{7}^{100}}=\left( 10000\times k \right)-\dfrac{50!}{47!\left( 50-47 \right)!}{{\left( 50 \right)}^{3}}+\dfrac{50!}{48!\left( 50-48 \right)!}{{\left( 50 \right)}^{2}}-\dfrac{50!}{49!\left( 50-49 \right)!}\left( 50 \right)+\dfrac{50!}{50!\left( 50-50 \right)!} \\
 & \Rightarrow {{7}^{100}}=\left( 10000\times k \right)-\dfrac{50\times 49\times 48\times 47!}{47!3!}{{\left( 50 \right)}^{3}}+\dfrac{50\times 49\times 48!}{48!2!}{{\left( 50 \right)}^{2}}-\dfrac{50\times 49!}{49!1!}\left( 50 \right)+1 \\
 & \Rightarrow {{7}^{100}}=\left( 10000\times k \right)-\dfrac{50\times 49\times 48}{3!}{{\left( 50 \right)}^{3}}+\dfrac{50\times 49}{2!}{{\left( 50 \right)}^{2}}-\dfrac{50}{1!}\left( 50 \right)+1 \\
\end{align}\]
Here, we can see that the term of \[{{50}^{3}}\] where we again get last four digits as zeros because the total multiplication gets \[{{50}^{4}}\]
Now, by rewriting the above equation by joining the first and second terms in the above equation we get
\[\Rightarrow {{7}^{100}}=\left( 10000\times p \right)+\dfrac{50\times 49}{2!}{{\left( 50 \right)}^{2}}-\dfrac{50}{1!}\left( 50 \right)+1\]
Now, by multiplying the remaining terms in above equation we get
\[\begin{align}
  & \Rightarrow {{7}^{100}}=\left( 10000\times p \right)+\left( 25\times 49\times 2500 \right)-\left( 2500 \right)+1 \\
 & \Rightarrow {{7}^{100}}=\left( 10000\times p \right)+3062500-2500+1 \\
 & \Rightarrow {{7}^{100}}=\left( 10000\times p \right)+3060001 \\
\end{align}\]
Now, we know that adding any number to zeros we get the same number.
Here, we can see that the last four digits in the first term are zeros.
So, adding the last four digits of the second term to the first term we get the same digits as in the second term.
Therefore we can conclude that the last four digits of \[{{7}^{100}}\] are 0001
So, option (d) is the correct answer.


Note:
 Students may do mistake in converting the given number in the form \[{{\left( x+1 \right)}^{n}}\] or \[{{\left( x-1 \right)}^{n}}\] where \[x\] is a multiple of 10.
Here we need the number \[x\] as the multiple of 10 because we can eliminate the terms that are having the last four digits as zeros.
Let us assume that the number \[x\] is not a multiple of 10 then, in order to eliminate the terms having the last four digits as zeros we need to expand each and every term in the binomial expansion. This gives a difficult procedure.
But, when we have \[x\] as the multiple of 10 we can easily eliminate the terms of having the last four digits as zeros because \[{{10}^{n}}\] is obtained by placing \[n\] zeros after 1.