Exercise 4.1.1 (de Polignac's formula)

Proof.

(a)

Using Theorem 4.1(6), the de Polignac’s formula (Theorem 4.2) gives $$\begin{array}{lll}\hfill {\nu }_p(n!)=\lfloor n∕p\rfloor +{\nu }_p(\lfloor n∕p\rfloor !).& & \hfill \text{(1)}\end{array}$$

Here

$$\begin{array}{ccccc}\lfloor 533∕2\rfloor =266,\hfill & \lfloor 266∕2\rfloor =133,\hfill & \lfloor 133∕2\rfloor =66,\hfill & \lfloor 66∕2\rfloor =33,\hfill & \lfloor 33∕2\rfloor =16,\hfill \\ \lfloor 16∕2\rfloor =8,\hfill & \lfloor 8∕2\rfloor =4,\hfill & \lfloor 4∕2\rfloor =2,\hfill & \lfloor 2∕2\rfloor =1.\hfill \end{array}$$

The sum of these results is

$${\nu }_2(533!)=529.$$

(b)

Similarly (see the Python program in the note below), $${\nu }_3(533!)=263.$$

(c)

By parts (a) and (b), the decomposition of $533!$ in primes begins with $533!=2^{529}\cdot 3^{263}\cdot \cdots$, thus $6^{263}\mid 533!$, but $6^{264}\nmid 533!$. Therefore the highest power of $6$ dividing $533!$ is $$\begin{array}{llll}\hfill \min <!--\; FUNCTION\; APPLICATION\; -->({\nu }_2(533!),{\nu }_3(533!))& ={\nu }_3(533!)& \hfill & \\ \hfill & =263.& \hfill & \end{array}$$

(d)

Since $533!=2\cdot 4^{264}\cdot 3^{263}\cdot m=8\cdot 12^{263}m$, where $m\wedge 12=1$, the highest power of $12$ dividing $533!$ is $263$.

(e)

As in part (c), the highest power of $70=2\cdot 5\cdot 7$ dividing $533!$ is $$\begin{array}{llll}\hfill \min <!--\; FUNCTION\; APPLICATION\; -->({\nu }_2(533!),{\nu }_5(533!),{\nu }_7(533!))& ={\nu }_7(533!)& \hfill & \\ \hfill & =87.& \hfill & \end{array}$$