Lucky Number 17!

Note first that for $n \ge 6$ we have that $6! | n!$ , and that $6! = 16 \times 45$ . Now as $17$ is prime, for $6 \le n \le 16$ we know that $17$ does not divide $n$ nor any (integral) power of $n$ . Thus by Fermat's Little Theorem , for $6 \le n \le 16$ we have that

$n^{n!} = (n^{n*(n - 1)*...*7})^{6!} = (n^{n*(n-1)*...*7*45})^{16} \equiv 1 \pmod{17}$ .

Then as $17 | 17^{17!}$ we see that $\displaystyle\sum_{n=6}^{17} (n^{n!} \pmod{17}) = \sum_{n=6}^{16} 1 = 11$ .

So now we need to evaluate $n^{n!} \pmod{17}$ for $1 \le n \le 5$ :

• $1^{1!} = 1 \equiv 1 \pmod{17}$

• $2^{2!} = 4 \equiv 4 \pmod{17}$

• $3^{3!} = (3^{3})^{2} = 27^{2} \equiv 10^{2} \pmod{17} \equiv 100 \pmod{17} \equiv 15 \pmod{17}$

• $4^{4!} = (4^{2})^{3*4} \equiv (-1)^{12} \pmod{17} \equiv 1 \pmod{17}$

• $5^{5!} = 5^{120} = 5^{7*16 + 8} \equiv 5^{8} \pmod{17} \equiv (5^{2})^{4} \pmod{17}$

$\equiv 8^{4} \pmod{17} \equiv 64^{2} \pmod{17} \equiv (-4)^{2} \pmod{17} \equiv 16 \pmod{17}$ ,

where another application of FLT was used in the third step of this last calculation. Thus

$\displaystyle\sum_{n=1}^{17} (n^{n!} \pmod{17}) = 1 + 4 + 15 + 1 + 16 + 11 = 48 \equiv \boxed{14} \pmod{17}$ .