First slide

Permutations

Question

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

Moderate

A

2
B

n
C

2n
D

none of these

Solution

Note that $d |n \Leftrightarrow \frac{n}{d}| n$
Now, $\sum_{j = 1}^{k} \frac{1}{d_{j}} = \sum_{j = 1}^{k} \frac{d_{j}}{n} = \frac{1}{n} \sum_{j = 1}^{k} d_{j}$
But divisors of $n = 2^{100} 3^{2}$ are
$1, 2, \dots, 2^{100}, 3, (3) (2), \dots, 3 (2^{100}), 3^{2} (3^{2}) (2), \dots$
Thus, $\sum_{j = 1}^{k} d_{j} = (1 + 2 + \dots + 2^{100}) (1 + 3 + 3^{2})$
$= (\frac{2^{101} - 1}{2 - 1}) (13) = 13 (2^{101} - 1)$

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