JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Permutations and Combinations

Since $91=7\times 13$, having $\gcd (91,n)>1$ means $n$ is a multiple of $7$ or of $13$. Between $100$ and $200$ the multiples of $7$ run $105,112,\dots ,196$, fourteen terms summing to $\frac{14}{2}(105+196)=2107$; the multiples of $13$ run $104,117,\dots ,195$, eight terms summing to $\frac{8}{2}(104+195)=1196$; and the only common multiple (of $91$) is $182$. By inclusion-exclusion the total is $2107+1196-182=3121$.