Find the greatest positive integer $x$ such that $23^{6+x}$ divides $2000!$

Question

I'm currently reading Andreescu and Andrica's Number Theory: Structures, examples and problems. Problem 1.1.7 states : Find the greatest positive integer $x$ such that $23^{6+x}$ divides $2000!$.

The solution given is :

The number $23$ is prime and divides every $23^{rd}$ number. In all, there are $\lfloor \frac{2000}{23} \rfloor$ = $86$ numbers from $1$ to $2000$ that are divisible by $23$. Among those $86$ numbers, three of them, namely $23, 2 · 23$ and $3 · 23^2$ are divisible by $23^3$. Hence $23^{89} | 2000!$ and $x = 89 − 6 = 83$.

I really don't understand how $23, 2 · 23$ and $3 · 23^2$ are divisible by $23^{3}$ and I hope someone here can shed light on this step in particular, and the overall solution.

EDIT: It appears to have been an error by the authors.

possible duplicate of Highest power of a prime $p$ dividing $N!$ — lab bhattacharjee, May 29 '15 at 19:01
@labbhattacharjee: Not a duplicate. This question asks about a specific claim in a specific solution -- giving a different (or general) solution to the problem type would not be an answer to this question. — hmakholm left over Monica, May 29 '15 at 19:04

score 5 · Accepted Answer · answered May 29 '15 at 19:02

5

It must be a typo. The list should read "$23^2$, $2\cdot 23^2$ and $3\cdot 23^2$ are divisible by $23^2$".

answered May 29 '15 at 19:02

hmakholm left over Monica

286,031

Thank you, this makes perfect sense. – Munyari May 29 '15 at 19:18

Find the greatest positive integer $x$ such that $23^{6+x}$ divides $2000!$

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