Prime Factors of 13 Factorial
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Example of Factorial
The prime decomposition of $13!$ is given as:
- $13! = 2^{10} \times 3^5 \times 5^2 \times 7 \times 11 \times 13$
Proof
For each prime factor $p$ of $13!$, let $a_p$ be the integer such that:
- $p^{a_p} \divides 13!$
- $p^{a_p + 1} \nmid 13!$
Taking the prime factors in turn:
| \(\ds a_2\) | \(=\) | \(\ds \sum_{k \mathop > 0} \floor {\frac {13} {2^k} }\) | De Polignac's Formula | |||||||||||
| \(\ds \) | \(=\) | \(\ds \floor {\frac {13} 2} + \floor {\frac {13} 4} + \floor {\frac {13} 8}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 6 + 3 + 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 10\) |
| \(\ds a_3\) | \(=\) | \(\ds \sum_{k \mathop > 0} \floor {\frac {13} {3^k} }\) | De Polignac's Formula | |||||||||||
| \(\ds \) | \(=\) | \(\ds \floor {\frac {13} 3} + \floor {\frac {13} 9}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 4 + 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 5\) |
| \(\ds a_5\) | \(=\) | \(\ds \sum_{k \mathop > 0} \floor {\frac {13} {5^k} }\) | De Polignac's Formula | |||||||||||
| \(\ds \) | \(=\) | \(\ds \floor {\frac {13} 5}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2\) |
| \(\ds a_7\) | \(=\) | \(\ds \sum_{k \mathop > 0} \floor {\frac {13} {7^k} }\) | De Polignac's Formula | |||||||||||
| \(\ds \) | \(=\) | \(\ds \floor {\frac {13} 7}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
Similarly:
- $a_{11} = 1$
- $a_{13} = 1$
Hence the result.
$\blacksquare$