QJEEM19Ma9d25: JEE Main 2019 Mathematics - Class 11 JEEM19Ma9d25 Solution: The sum of all natural numbers 'n' such that 100 1 is :

$\because$ 91 = 13 $\times$ 7

So the required numbers are either divisible by 7 or 13.

S_A = sum of numbers between 100 and 200 which are divisible by 7.

$\Rightarrow$ S_A = 105 + 112 + ..... + 196

S_A = ${{14} \over 2}\left[ {105 + 196} \right]$ = 2107

S_B = Sum of numbers between 100 and 200 which are divisible by 13.

S_B = 104 + 117 + ....... + 195

S_B = ${8 \over 2}\left[ {104 + 195} \right]$ = 1196

S_C = Sum of numbers between 100 and 200 which are divisible by 7 and 13.

S_C = 182

Sum of numbers divisible by 7 or 13 = Sum of no. divisible by 7 + sum of the no. divisible by 13 – Sum of the numbers divisible by 7 and 13

= 2107 + 1196 - 182

= 3121