Competition problem — Number theory

Library / /289 of 520

Number theory Difficulty 6.2 National olympiad Prove it

Lemma 6.1. Let $p$ be prime and $a$ a positive integer. Then
$\sigma\left(p^{a}\right)=\left(1+p+p^{2}+\cdots+p^{a}\right)=\frac{p^{a+1}-1}{p-1}$
and
$\tau\left(p^{a}\right)=a+1$

Solution

Proof. The divisors of $p^{a}$ are $1, p, p^{2}, \ldots, p^{a-1}, p^{a}$ . Consequently, $p^{a}$ has exactly $a+1$ divisors, so that $\tau\left(p^{a}\right)=a+1$ . Also, we note that $\sigma\left(p^{a}\right)=1+p+p_{2}+\cdots+p^{a-1}+p^{a}=\frac{p^{a+1}-1}{p-1}$ , where we have used T

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.