Summation of logarithm of divisors

$\sum_{d|10^6} \log_{10} d = \sum_{d_1|2^6,d_2|5^6} \log_{10} (d_1 \cdot d_2)$

$=\sum_{d_1|2^6,d_2|5^6} \log_{10} d_1 + \log_{10} d_2 = \sum_{d_1|2^6,d_2|5^6} \log_{10} d_1 + \sum_{d_1|2^6,d_2|5^6}\log_{10} d_2$

Note that $2^6$ has $7$ divisors.

$7 \cdot \sum_{t=0}^{6} \log_{10}2^t + 7 \cdot \sum_{t=0}^{6} \log_{10}5^t$

$7 \cdot (1+\dots+6)\cdot \log_{10} 2 + 7 \cdot (1+\dots+6)\cdot \log_{10} 5 = 7 \cdot 21 \cdot (\log_{10} 2 +\log_{10} 5 ) = 147$

$10^6 = 2^6 5^6$ has $(6+1)(6+1)=49$ divisors $d$ . Number the divisors from the smallest to the largest. That is $d_1 = 1, d_2 = 2, d_3 = 4, \cdots, d_{49}=10^6$ . Then

$\begin{aligned} \sum_{d \mid a} \log_{10} d & = \sum_{n=1}^{49} \log_{10} d_n \\ & = \sum_{n=1}^{24} \log_{10} d_n + \log_{10} d_{25} + \blue{\sum_{n=26}^{49} \log_{10} d_n} & \small \blue{\text{Note that }d_{49} = \frac {10^6}{d_1}, d_{48} = \frac {10^6}{d_2}, \cdots} \\ & = \sum_{n=1}^{24} \log_{10} d_n + \log_{10} 10^3 + \blue{\sum_{n=26}^{49} \log_{10} \frac {10^6}{d_{50-n}}} \\ & = \sum_{n=1}^{24} \log_{10} d_n + 3 + \sum_{n=1}^{24} \log_{10} \frac {10^6}{d_1} \\ & = \sum_{n=1}^{24} \left(\log_{10} d_n + \log_{10} \frac {10^6}{d_1} \right) + 3 \\ & = 3 + \sum_{n=1}^{24} \left(\log_{10} d_n + \log_{10} 10^6 - \log_{10} d_1 \right) \\ & = 3 + \sum_{n=1}^{24} 6 = 3 + 24(6) = \boxed {147} \end{aligned}$