If $$n=210032$$ and $$d1$$,$$d2$$,…dk is the set of all divisors of n, then ∑$$j=1k$$ $$1dj$$ equals

Correct option is 4none of these

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If $n = 2^{100} 3^{2}$ and $\{d_{1}, d_{2}, \dots d_{k}\}$ is the set of all divisors of $n$ , then $\sum_{j = 1}^{k} \frac{1}{d_{j}}$ equals

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Correct option is D

Note that dn⇔ndnNow, ∑j=1k 1dj=∑j=1k djn=1n∑j=1k djBut divisors of n=210032 are1,2,⋯,2100,3,(3)(2),⋯,32100,3232(2), ⋯Thus, ∑j=1k dj=1+2+⋯+21001+3+32=2101−12−1(13)=132101−1

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If n=210032 and d1,d2,…dk is the set of all divisors of n, then ∑j=1k 1dj equals

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If $n = 2^{100} 3^{2}$ and $\{d_{1}, d_{2}, \dots d_{k}\}$ is the set of all divisors of $n$ , then $\sum_{j = 1}^{k} \frac{1}{d_{j}}$ equals