TITLE: How would I find a number where $\sum_{d\mid n}d > 100n$?
QUESTION [5 upvotes]: How would I find an $n$ such that
$$
\sum_{d\mid n} d > 100 n
$$
I have absolutely no idea as to how to begin this question.

REPLY [7 votes]: From a  theorem of Robin we have an unconditional result (1984) that says that for $n \geq 13,$ we have the bound
$$ \frac{\sigma(n)}{n} < \; e^\gamma \log \log n + \frac{0.64821364942...}{\log \log n},$$ with the constant in the numerator giving equality for $n=12.$ Note
$$  e^\gamma  = 1.7810724179901979852365\ldots  $$ Let's see, the Riemann Hypothesis is equivalent to the statement that, for $n \geq 5041 = 1 + 7!,$ the number $0.64821364942...$ can be replaced by $0.$ 
In order for your ratio to reach 100, you need 
$$   n > e^{\left( e^{56.14594836} \right)}  $$
which is a very, very big number.  See http://en.wikipedia.org/wiki/Colossally_abundant_number and, for that matter, my answer at https://mathoverflow.net/questions/79927/which-n-maximize-gn-frac-sigmann-log-log-n 
EEEDDDDIIITTT: The smallest number for which your ratio is at least 100 is called a "superabundant number" by Alaoglu and Erdos (1944). The way I like to say it is that a superabundant number $n$ sets a new world record for $\sigma(n)/n.$ On journal numbered page 450 we find Theorem 1, such numbers have nonincreasing exponents, and Theorem 2, all the exponents can be found within an error of $\pm 1$ by finding the exponent of 2. They give more theorems. The upshot is that superabundant numbers have factorizations very similar to those of a subsequence called the Colossally Abundant Numbers. Given a positive real number they call $\epsilon,$ the exponent of a prime $q$ is given in Theorem 10 on journal page 455. Finally, on pages 468-469, they display all superabundant numbers up to $10^{18}.$ Those numbers in this table that are also colossally
abundant are labelled. The largest of these, with $\sigma(n)/n \approx 6.407,$ can be found with $\epsilon = 0.007.$
EDIT: The software still needs work, but I found a small number with your ratio exceeding 10:
n   3087412123011300495851995214301626485649741980169578200745289299789026306722292583565231315659291930544327540696937285310080000
sigma 30941752989588328135737108482280837416279110376522697522825909747010685327091094608601433751156507323392138936320000000000000000
mpf  10.0219
This number $n$ is a CA for $\epsilon = 0.00064.$
Why not, the number displayed is
$$ n = 2^{10} \, 3^6 \, 5^4 \, 7^3 \, 11^2 \, 13^2 \, 17^2 \, 19^2 \cdot 23 \cdot 29 \cdots 271 \cdot 277  $$
I found the first million superabundant numbers at a link at http://oeis.org/A004394 and excerpted just a few lines below. It turns out I did not do too badly with ratio 10. It is the 853'd SA number that first gives $\sigma(n) \geq 10n.$ I gave the 860th SA number. Note that the 1,000,000th SA number still has $\sigma(n)/n \approx 22.04,$ so 100 requires a really big SA number, probably so big that one should just be satisfied with the first CA number that succeeds. I threw in the first number with ratio at least 20. I also put in the first CA number with ratio at least 20, the relevant $\epsilon = 0.00000118087 =  1.18087 \cdot 10^{-6}$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
// This file contains information about the first 1000000 superabundant
// numbers (SA numbers).  For each SA number n, we give its abundance,
// which is Sigma(n)/n, its base-10 logarithm, and its factorization.
// In column "*", a "C" indicates that the number is also a colossally
// abundant number  The factorization is given in a very compact form.
// For example, {13,5,0,2} means 13 * 11 * 7 * 5^2 * 3^2 * 2^4.
//
// Created by Tony D. Noe, noe@sspectra.com on 15-Oct-2009.
// Algorithm developed with help from Devin Kilminster.
//
// Corrected 30-Oct-2009: some SA numbers were erroneously marked with a "C".
//
// position       abundance                log10     *   factorization
//
         1   1.0000000000000000         0.0000000000 S {0}
         2   1.5000000000000000         0.3010299957 C {2}
         3   1.7500000000000000         0.6020599913 S {0,2}
         4   2.0000000000000000         0.7781512504 C {3}
         5   2.3333333333333333         1.0791812460 C {3,2}
         6   2.5000000000000000         1.3802112417 S {3,0,2}
         7   2.5277777777777778         1.5563025008 S {0,3}
         8   2.5833333333333333         1.6812412374 S {3,0,0,2}
         9   2.8000000000000000         1.7781512504 C {5,2}
        10   3.0000000000000000         2.0791812460 C {5,0,2}

       100   6.2948845987914689        16.5059308532 S {37,5,0,3,0,2}

       853  10.0039459446460710       125.4088426721 S {271,23,7,5,0,3,0,0,0,2}

       860  10.0219056467943640       126.4895946051 C {277,19,7,5,0,3,0,0,0,2}

      1000  10.3332076452686825       149.9416130802 S {337,23,7,5,0,3,0,0,2}

     10000  14.3329541785168048      1378.4010046121 S {3163,79,19,7,5,0,0,0,3,0,0,0,2}

    100000  18.2270550779625505     12146.0718137217 S {27893,229,41,17,0,7,5,0,0,3,0,0,0,0,0,0,0,2}

    297146  20.0000073661334977     32800.4952412562 S {75401,373,53,19,11,7,0,5,0,0,0,3,0,0,0,0,0,0,2}

    297210  20.0002475182905199     32804.8955086267 C {75403,373,53,19,11,7,0,5,0,0,3,0,0,0,0,0,0,0,2}

   1000000  22.0419355366430523    103082.8167397490 S {237173,673,83,31,13,0,7,5,0,0,0,0,3,0,0,0,0,0,0,2}

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Sunday, 7 April 2013: It took a while, but I found the first Colossally Abundant number with $\sigma(n) / n \geq 30. $ It is easy to show that such numbers are superabundant. 
$$ 2^{27}  3^{17} 5^{11} 7^9 11^7 13^7 17^6 19^6 23^5 \cdots 37^5 41^4 \cdots 89^4 97^3 \cdots 383^3 389^2 \cdots 6301^2 \cdot 6311 \cdots 20663801.    $$
This number is about 
$$  10^{8974734.77703953}  $$
The Alaoglu-Erdos parameter is $$ \epsilon = 0.000000002873077 = 2.873077 \cdot 10^{-9}.   $$
Oh, $$ \sigma(n)/n \approx 30.00000029489  $$