Q. The digit at the unit place in the number $19^{2005} + 11^{2005} - 9^{2005}\ is$
A) 2
B) 1
C) 0
D) 8
Explanation
(b)
$(19)^{2005} + (11)^{2005} - (9)^{2005}$
$= (10 + 9)^{2005} + (10 + 1)^{2005} - (9)^{2005}$
$= (9^{2005} + \ ^{2005}C_{1}(9)^{2004} \times 10 +$…..)$+ \left( \ ^{2005}C_{0} + \ ^{2005}C_{1}10 + ... \right) - (9)^{2005}$
$= (\ ^{2005}C_{1}9^{2005} \times 10 +$multipale of 10)+(1+multipal 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.