Run on the ceiling

First of all We observe that,

If a,b,c are +ve integers & they satisfy $10^a < b < 10^c$ then :

$a < log_{10}b < c$ . ............................................................ (1)

Now 1729 = 7.13.19 & Number of divisors of 1729 is (1+1)(1+1)(1+1) i.e 8 including 1 & 1729 .

The 6 factors can be easily listed out except the number and 1 as follows , 7,13,19,91,133,247 .

Observe that,

[ $10^1 < 13,19,91 < 10^2$ ] & [ $10^2 < 133,247 < 10^3$ ]

So by (1) we can say that

[ $1 < \log_{10}13 , \log_{10}19 , \log_{10}91 < 2$ ] & [ $2 < \log_{10}133 , \log_{10}247 < 3$ ]

$\large \displaystyle \sum_{k | 1729}^{} k\lceil\log_{10} k\rceil = ?$

= $\large \displaystyle 1\lceil\log_{10}1\rceil + 7\lceil\log_{10}7\rceil + 13\lceil\log_{10}13\rceil + 19\lceil\log_{10}19\rceil + 91\lceil\log_{10}91\rceil + 133\lceil\log_{10}133\rceil + 247\lceil\log_{10}247\rceil + 1729\lceil\log_{10}1729\rceil$

since the $\log_{10}7$ lies between 0 & 1 it is 1, $\log_{10}13, \log_{10}19, \log_{10}91$ lie between 1 & 2 so they will equal 2 when subjected to the ceiling.

Similiarly the $\log_{10}133, \log_{10}247$ will equal 3 as the lie between 2 & 3 .

= $0 + 7(1) + (13+19+91)2 + (133+247)3 + 1729.4$

=8309 .