Question

The digit at the unit place in the number ${19}^{2005}+{11}^{2005}-9^{2005}$ is$-$

• A. $2$
• B. $1$
• C. $0$
• D. $9$

Answer 1

Answer: (B)

$1$

Rearranging the terms we get
$(1+10{)}^{2005}-(10-1{)}^{2005}+(20-1{)}^{2005}$
$=\left[\right(1+10{)}^{2005}+(1-10{)}^{2005}]-(1-20{)}^{2005}$
In the first bracket, all the even terms will get eliminated.
Hence we are left with
$2[1+{}^{2005}{C}_2{10}^2+{}^{2005}{C}_4{10}^4...{}^{2005}{C}_{2004}{10}^{2004}]-(1-{}^{2005}{C}_1{20}^1+{}^{2005}{C}_2{20}^2-{}^{2005}{C}_3{20}^3...{}^{}{}^{2005}{C}_{2005}{20}^{2005})$
$=2(1+100k)-(1+20m)$ ...(where m and k are positive integers).
$=(200k-20m)+(2-1)$
Hence units digits is given by $2-1=1$