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IMO 2J, Independent Study 1

Number theory Difficulty 8.3 Shortlist Prove it Taiwan

For positive integers $n$ and $k \ge 2$ define $E_k(n)$ as the greatest exponent $r$ such that $k^r$ divides $n!$ . Prove that there are infinitely many $n$ such that $E_{10}(n) > E_9(n)$ and infinitely many $m$ such that $E_{10}(m) < E_9(m)$ .

Solution

Set $n = 5^{2l-1}$ , we have
$E_{10}(n) = v_5(n!) = 5^{2l-2} + 5^{2l-3} + \dots + 5 + 1 = \frac{5^{2l-1}-1}{4} = \frac{n-1}{4}.$
Since $n \equiv 2 \pmod 4$ , so
$E_9(n) = \frac{1}{2}v_3(n!) < \frac{1}{2}\left(\frac{n-2}{3} + \frac{n}{3^2} + \dots\right) = \frac{n}{4} - \frac{1}{3}.$
Thus $E_9(n) < E_{10}(n)$ .

Similarly, set $m = 3^{4l-2}$ . Then we have
$E_9(m) = \frac{1}{2}v_3(m!) = \frac{m-1}{4}$
and since $m \equiv 4 \pmod 5$ ,
$E_{10}(m) < \frac{m-4}{5} + \frac{m}{5^2} + \dots = \frac{m}{4} - \frac{4}{5}.$
We have $E_{10}(m) < E_9(m)$

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from en; metadata (topic, difficulty) added by this project.