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. $(1,6)$
• B. $(6,1)$
• C. $(8,1)$
• D. $(1,8)$

Answer 1

Answer: (B)

$(6,1)$

We can write $(11{)}^{2016}=(10+1{)}^{2016}$

$={}^{2016}{C}_0{10}^{2016}+......+{}^{2016}{C}_{2014}{10}^2+{}^{2016}{C}_{2015}10+{}^{2016}{C}_{2016}$ .... Using binomial expansion

Since, ${10}^3=100$ is appearing every term so take $1000$ common from all the terms and we get

$(11{)}^{2016}=(10+1{)}^{2016}$

$=1000\lambda +203112000+20160+1$, where $\lambda ={}^{2016}{C}_0{10}^{2013}+{}^{2016}{C}_1{10}^{2012}+....+{}^{2016}{C}_{2013}$

$=1000\lambda +203,132,161$

The ten's digit is $6$, so $x=6$

Hundred's digit is $1$, so $y=1$

Hence, $(x,y)=(6,1)$