Q. The digit at the unit place in the number $19^{2005} + 11^{2005} - 9^{2005}$ is
A) 2
B) 1
C) O
D) 8
Explanation
(b)
$19^{2005} + 11^{2005} - 9^{2005}$
${= (10 + 9)}^{2005} + {(10 + 1)}^{2005} - ({9)}^{2005}$
$= \left( 9^{2005} + \ ^{2005}C_{1}(9)^{2004} \times 10 + ... \right) + \left( \ ^{2005}C_{0} + \ ^{2005}C_{1}10 + ... \right) - (9)^{2005}$
$= \left( \ ^{2005}C_{1}9^{2004} \times 10 + \ multiple\ of\ 10 \right) + (1 + \ multiple\ of\ 10)$
$\therefore$ Unit digit = 1
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.