Q. A three digit number $n$ is such that the last two digits of it are equal and differ from the first. The number of such $n$’s is
A) 64
B) 72
C) 81
D) 900
Explanation
(c)
If the last two digits are 0, 0 then in 1st digit any of the numbers expect 0 $ie$, 9 numbers
If the last two digits are 1, 1, then in 1st digit any of the numbers expect 0 and 1 $ie$, 8 numbers
$\therefore$ The total number of numbers$= 9 + 8 \times 9 = 81$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.