QJEEM21Mad0ed: JEE Main 2021 Mathematics - Class 11 JEEM21Mad0ed Solution: The sum of all 3-digit numbers less than or equal to 500, that…

209, 220, 231, ..........., 495

Sum = ${{27} \over 2}$ (209 + 495) = 9504

Number containing 1 at unit place $$\begin{array}{ccc}\underset{―}{2}& \underset{―}{3}& \underset{―}{1}\\ \underset{―}{3}& \underset{―}{4}& \underset{―}{1}\\ \underset{―}{4}& \underset{―}{5}& \underset{―}{1}\end{array}$$

Number containing 1 at 10^th place $$\begin{array}{ccc}\underset{―}{3}& \underset{―}{1}& \underset{―}{9}\\ \underset{―}{4}& \underset{―}{1}& \underset{―}{8}\end{array}$$

Required = 9504 $-$ (231 + 341 + 451 + 319 + 418)

= 7744