JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 5)
The sum of all natural numbers 'n' such that
100 < n < 200 and H.C.F. (91, n) > 1 is :
3221
3121
3203
3303
Explanation
$$ \because $$ 91 = 13 $$ \times $$ 7
So the required numbers are either divisible by 7 or 13.
SA = sum of numbers between 100 and 200 which are divisible by 7.
$$ \Rightarrow $$ SA = 105 + 112 + ..... + 196
SA = $${{14} \over 2}\left[ {105 + 196} \right]$$ = 2107
SB = Sum of numbers between 100 and 200 which are divisible by 13.
SB = 104 + 117 + ....... + 195
SB = $${8 \over 2}\left[ {104 + 195} \right]$$ = 1196
SC = Sum of numbers between 100 and 200 which are divisible by 7 and 13.
SC = 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
So the required numbers are either divisible by 7 or 13.
SA = sum of numbers between 100 and 200 which are divisible by 7.
$$ \Rightarrow $$ SA = 105 + 112 + ..... + 196
SA = $${{14} \over 2}\left[ {105 + 196} \right]$$ = 2107
SB = Sum of numbers between 100 and 200 which are divisible by 13.
SB = 104 + 117 + ....... + 195
SB = $${8 \over 2}\left[ {104 + 195} \right]$$ = 1196
SC = Sum of numbers between 100 and 200 which are divisible by 7 and 13.
SC = 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
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