JEE PYQ: Permutation And Combination Question 32
Question 32 - 2019 (08 Apr Shift 1)
The sum of all natural numbers ‘$n$’ such that $100 < n < 200$ and H.C.F. $(91, n) > 1$ is:
(1) 3203
(2) 3303
(3) 3221
(4) 3121
Show Answer
Answer: (4)
Solution
$91 = 7 \times 13$. Numbers divisible by 7: $105, 112, \ldots, 196$. Sum $= 2107$. Numbers divisible by 13: $104, 117, \ldots, 195$. Sum $= 1196$. Numbers divisible by 91: $182$. Sum $= 182$. By inclusion-exclusion: $2107 + 1196 - 182 = 3121$.
BETA
