# Sum of divisors and prime numbers, short proof

Let $p_i$ denote the $i^{th}$ prime number.

Find the smallest positive integer $k$ such that the product $n = p_1 \cdot p_2 \cdots p_k$ satisfies $\sigma(n) > 3n$.

Is there any positive integer $m < n$ satisfying $\sigma(m) > 3m$?

Taken From (Rosen), at the end of the chapter under challenge problems

-
$\sigma$ is sum of divisors I presume? Also, this sounds like homework. Is it? – Aryabhata Sep 4 '10 at 20:12
It's a question from my number theory book (Rosen), at the end of the chapter under challenge problems. Thank you for editing, and yes, $\sigma$ is sum of divisors. – gurk Sep 4 '10 at 20:15
@gurk: You'd better add back the source of the question (Rosen). It is discouraged to post text without citing it. – kennytm Sep 4 '10 at 20:21
If $\gcd(m,n)=1$, then $\sigma(mn)=\sigma(m)\sigma(n)$; and $\sigma(p)$ is easy to compute when $p$ is a prime. So you have an explicit expression for $\sigma(n)$, at least, which should solve the first part of the problem. Then think about the prime factorization of any $m\lt n$. – Arturo Magidin Sep 4 '10 at 20:31
Thank you Arturo, I think I understand how to proceed now. Thanks for the push in the right direction, greatly appreciate it. – gurk Sep 4 '10 at 20:38

The first $k$ for which $n = \prod_{i \leqslant k} p_k$ and $\sigma (n) > 3n$ is $6$, here $n = 30030$. Also, $m = 240 < 30030$ is the least integer with $\sigma (m) > 3m$.