JEE Main 2019 — Sequences And Series Question with Solution
From: JEE Main 2019 (Online) 8th April Morning Slot
Question
The sum of all natural numbers 'n' such that
100 < n < 200 and H.C.F. (91, n) > 1 is :
Choose an option
Show full solutionCorrect option: B
Correct answer
B3121
Step-by-step explanation
91 = 13 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.
SA = 105 + 112 + ..... + 196
SA = = 2107
SB = Sum of numbers between 100 and 200 which are divisible by 13.
SB = 104 + 117 + ....... + 195
SB = = 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.
SA = 105 + 112 + ..... + 196
SA = = 2107
SB = Sum of numbers between 100 and 200 which are divisible by 13.
SB = 104 + 117 + ....... + 195
SB = = 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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