Question

If the digits at ten’s and hundred’s places in ${\left( {11} \right)^{2016}}$ are x and y respectively, then the ordered pair (x, y) is equal to-
A.$\left( {1,6} \right)$
B.$\left( {6,1} \right)$
C.$\left( {8,1} \right)$
D.$\left( {1,8} \right)$

Answer 1

Hint: We can use expression of binomial theorem which is given as-
\[ \Rightarrow {\left( {x + a} \right)^n} = {}^n{C_0}{x^n}{a^0} + {}^n{C_1}{x^{n - 1}}{a^1} + ... + {}^n{C_n}{x^{n - n}}{a^n}\] Where,${}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$ and n is positive integral.
Take out the common terms in the expression and solve to find the value of x and y.

Complete step-by-step answer:
Given, the ten’s digit in ${\left( {11} \right)^{2016}}$=x
And the one’s digit in ${\left( {11} \right)^{2016}}$=y
We have to find the value of ordered pair (x, y)
We can write \[{\left( {11} \right)^{2016}} = {\left( {10 + 1} \right)^{2016}}\] -- (i)
Now using expression of binomial theorem-
\[ \Rightarrow {\left( {x + a} \right)^n} = {}^n{C_0}{x^n}{a^0} + {}^n{C_1}{x^{n - 1}}{a^1} + ... + {}^n{C_n}{x^{n - n}}{a^n}\] Where,${}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$ and n is positive integral.
On putting the values in formula we get,
\[ \Rightarrow {\left( {11} \right)^{2016}} = {\left( {10 + 1} \right)^{2016}} = {}^{2016}{C_0}{.10^{2016}} + {}^{2016}{C_1}{.10^{2016 - 1}}{1^1} + ... + {}^{2016}{C_{2014}}{.10^{2016 - 2014}}{1^{2014}} + {}^{2016}{C_{2015}}{.10^{2016 - 2015}}{1^{2015}} + {}^{2016}{C_{2016}}{.10^{2016 - 2016}}{1^{2016}}\] On simplifying we get,
\[ \Rightarrow {\left( {11} \right)^{2016}} = {\left( {10 + 1} \right)^{2016}} = {}^{2016}{C_0}{.10^{2016}} + {}^{2016}{C_1}{.10^{2015}} + ... + {}^{2016}{C_{2014}}{.10^2} + {}^{2016}{C_{2015}}.10 + {}^{2016}{C_{2016}}\]
We know that ${}^n{C_0} = {}^n{C_n} = 1$
Then on applying this in the above equation, we get
\[ \Rightarrow {\left( {11} \right)^{2016}} = {\left( {10 + 1} \right)^{2016}} = {10^{2016}} + {}^{2016}{C_1}{.10^{2015}} + ... + {}^{2016}{C_{2014}}{.10^2} + {}^{2016}{C_{2015}}.10 + 1\]
On using formula ${}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$
We get,
\[ \Rightarrow {\left( {11} \right)^{2016}} = {\left( {10 + 1} \right)^{2016}} = {10^{2016}} + \dfrac{{2016!}}{{2015!}}{.10^{2015}} + ... + \dfrac{{2016!}}{{2014!2}}{.10^2} + \dfrac{{2016!}}{{2015!}}.10 + 1\]
On taking ${10^3}$ common from every term except last three term (as it is not common in them) we get,
\[ \Rightarrow {\left( {11} \right)^{2016}} = {\left( {10 + 1} \right)^{2016}} = {10^3}\lambda + \left( {\dfrac{{2016 \times 2015}}{2}} \right) \times {10^2} + \left( {2016 \times 10} \right) + 1\] -- (ii)
Where $\lambda = {10^{2013}} + \dfrac{{2016!}}{{2015!}}{.10^{2012}} + ... + \dfrac{{2016!}}{{2013!3!}}$
On solving eq. (i) further, we get
\[ \Rightarrow {\left( {11} \right)^{2016}} = {\left( {10 + 1} \right)^{2016}} = {10^3}\lambda + \left( {2015 \times 1008} \right) \times 100 + 20160 + 1\]
On multiplication we get,
\[ \Rightarrow {\left( {11} \right)^{2016}} = {\left( {10 + 1} \right)^{2016}} = {10^3}\lambda + 2,031,120 \times 100 + 20161\]
\[ \Rightarrow {\left( {11} \right)^{2016}} = {\left( {10 + 1} \right)^{2016}} = {10^3}\lambda + 203,112,000 + 20161\]
On adding the last two terms we get,
\[ \Rightarrow {\left( {11} \right)^{2016}} = {\left( {10 + 1} \right)^{2016}} = {10^3}\lambda + 203,122,161\] -- (iii)
Here since the first term is multiple of ${10^3}$ so this means that whatever the value of $\lambda $ be, it will not affect the last three digits of the second term when added to it.
So the hundred’s ten’s and one’s digits are fixed. And it was given that the ten’s digit is x and one’s digit is y, then
$ \Rightarrow x = 6$ and
$ \Rightarrow y = 1$
So the ordered pair (x, y) will be $(6,1)$
Hence, the correct answer is ‘B’.

Note: In this question if we exchange the place of $10$ and $1$ in eq. (i), and then apply the binomial theorem and solve in the same method as above, the terms of the equation (iii) will also be exchanged. The answer we will get will be the same.