Prime Factors of 20 Factorial

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Example of Factorial

The prime decomposition of $20!$ is given as:

$20! = 2^{18} \times 3^8 \times 5^4 \times 7^2 \times 11 \times 13 \times 17 \times 19$


Proof

For each prime factor $p$ of $20!$, let $a_p$ be the integer such that:

$p^{a_p} \divides 20!$
$p^{a_p + 1} \nmid 20!$


Taking the prime factors in turn:

\(\ds a_2\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {20} {2^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {20} 2} + \floor {\frac {20} 4} + \floor {\frac {20} 8 } + \floor {\frac {20} {16} }\)
\(\ds \) \(=\) \(\ds 10 + 5 + 2 + 1\)
\(\ds \) \(=\) \(\ds 18\)


\(\ds a_3\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {20} {3^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {20} 3} + \floor {\frac {20} 9}\)
\(\ds \) \(=\) \(\ds 6 + 2\)
\(\ds \) \(=\) \(\ds 8\)


\(\ds a_5\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {20} {5^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {20} 5}\)
\(\ds \) \(=\) \(\ds 4\)


\(\ds a_7\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {20} {7^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {20} 7}\)
\(\ds \) \(=\) \(\ds 2\)

Similarly:

$a_{11} = 1$
$a_{13} = 1$
$a_{17} = 1$
$a_{19} = 1$

Hence the result.

$\blacksquare$


Sources