1. **Problem Statement:** We have $p=3$ and the set $$ A = \{ n \in \mathbb{Z} \mid 1 \le n \le 243, \ \gcd(n,3)=1 \}. $$ For each $n \in A$, define $$ f(n) = \nu_3\left(n^{81} - 1\right), $$ where $\nu_3(m)$ is the 3-adic valuation of $m$ (the exponent of the highest power of 3 dividing $m$). We want to find $$ S = \sum_{n \in A} f(n). $$ 2. **Understanding the problem:** - $A$ consists of integers from 1 to 243 that are coprime to 3. - $f(n)$ measures how many times 3 divides $n^{81} - 1$. - We sum $f(n)$ over all such $n$. 3. **Key facts and formulas:** - Since $p=3$, $|A|$ is the count of integers from 1 to 243 not divisible by 3. - $243 = 3^5$. - Euler's totient $\varphi(243) = 243 \times (1 - \frac{1}{3}) = 162$. - The group $(\mathbb{Z}/243\mathbb{Z})^*$ has order 162. - For $n$ coprime to 3, $n^{162} \equiv 1 \pmod{243}$ by Euler's theorem. 4. **Valuation of $n^{81} - 1$:** We want $\nu_3(n^{81} - 1)$ for $n$ with $\gcd(n,3)=1$. 5. **Using Lifting The Exponent Lemma (LTE):** For an odd prime $p$ and integers $a,b$ with $p \mid a-b$ but $p \nmid a,b$, and $k$ odd, $$ \nu_p(a^k - b^k) = \nu_p(a-b) + \nu_p(k). $$ Here, set $a=n$, $b=1$, $p=3$, and $k=81$ (which is $3^4$). Since $\gcd(n,3)=1$, $3 \nmid n-1$ unless $n \equiv 1 \pmod{3}$. 6. **Case 1: $n \equiv 1 \pmod{3}$** Then $3 \mid n-1$, so $\nu_3(n-1) \ge 1$. By LTE, $$ \nu_3(n^{81} - 1) = \nu_3(n-1) + \nu_3(81) = \nu_3(n-1) + 4. $$ 7. **Case 2: $n \not\equiv 1 \pmod{3}$** Then $3 \nmid n-1$, so $\nu_3(n-1) = 0$. LTE does not apply directly, but since $n^{162} \equiv 1 \pmod{243}$, and $81$ divides $162$, the order divides 162. Check $n^{81} - 1$ modulo 3: - If $n \equiv 2 \pmod{3}$, then $n^{81} \equiv 2^{81} \equiv 2 \pmod{3}$ (since $2^1=2$, $2^2=1$, cycle length 2). - So $n^{81} - 1 \equiv 2 - 1 = 1 \pmod{3}$, so $3 \nmid n^{81} - 1$. Hence $\nu_3(n^{81} - 1) = 0$ if $n \equiv 2 \pmod{3}$. 8. **Summary:** - If $n \equiv 1 \pmod{3}$, $f(n) = \nu_3(n-1) + 4$. - If $n \equiv 2 \pmod{3}$, $f(n) = 0$. 9. **Counting elements in $A$ by residue mod 3:** - $A$ excludes multiples of 3. - Numbers from 1 to 243: - $81$ numbers $\equiv 0 \pmod{3}$ (excluded) - $81$ numbers $\equiv 1 \pmod{3}$ - $81$ numbers $\equiv 2 \pmod{3}$ So $|A|=162$, with 81 in each residue class 1 and 2 mod 3. 10. **Sum $S$ becomes:** $$ S = \sum_{\substack{n=1 \,\le n \le 243 \\ n \equiv 1 \pmod{3}}} (\nu_3(n-1) + 4) + \sum_{\substack{n=1 \,\le n \le 243 \\ n \equiv 2 \pmod{3}}} 0 = \sum_{n \equiv 1 (3)} \nu_3(n-1) + 4 \times 81. $$ 11. **Rewrite $n$ with $n \equiv 1 \pmod{3}$:** Let $n = 3k + 1$ for $k=0,1,\ldots,80$. Then $$ n - 1 = 3k. $$ So $$ \nu_3(n-1) = \nu_3(3k) = 1 + \nu_3(k). $$ 12. **Sum over $k=0$ to $80$:** $$ \sum_{k=0}^{80} \nu_3(n-1) = \sum_{k=0}^{80} (1 + \nu_3(k)) = \sum_{k=0}^{80} 1 + \sum_{k=0}^{80} \nu_3(k) = 81 + \sum_{k=0}^{80} \nu_3(k). $$ Note $\nu_3(0)$ is infinite, but $k=0$ corresponds to $n=1$, so $n-1=0$ which is divisible by all powers of 3. Since $f(1) = \nu_3(0)$ is infinite, but the problem is well-defined, we interpret $\nu_3(0) = \infty$ meaning $f(1)$ is infinite. But $n=1$ is in $A$ and $f(1) = \nu_3(1^{81} - 1) = \nu_3(0) = \infty$. This suggests the problem expects us to exclude $n=1$ or treat it carefully. 13. **Check $f(1)$:** Since $1^{81} - 1 = 0$, $\nu_3(0)$ is infinite. So $f(1)$ is infinite, making $S$ infinite. 14. **Assuming the problem expects $n \neq 1$ or $f(1)=0$ (or ignoring $n=1$):** Sum over $k=1$ to $80$: $$ \sum_{k=1}^{80} \nu_3(k) + 81. $$ 15. **Calculate $\sum_{k=1}^{80} \nu_3(k)$:** Count how many numbers are divisible by 3, 9, 27, 81: - Multiples of 3: $\lfloor 80/3 \rfloor = 26$ - Multiples of 9: $\lfloor 80/9 \rfloor = 8$ - Multiples of 27: $\lfloor 80/27 \rfloor = 2$ - Multiples of 81: $\lfloor 80/81 \rfloor = 0$ Sum of valuations: $$ \sum_{k=1}^{80} \nu_3(k) = 26 + 8 + 2 + 0 = 36. $$ 16. **Therefore:** $$ \sum_{k=1}^{80} (1 + \nu_3(k)) = 80 \times 1 + 36 = 116. $$ 17. **Sum $S$ is:** $$ S = \sum_{k=1}^{80} (1 + \nu_3(k)) + 4 \times 81 = 116 + 324 = 440. $$ 18. **Final answer:** $$ \boxed{440}. $$ --- **Note:** If $n=1$ is included strictly, $f(1)$ is infinite, so the sum diverges. The problem likely expects to exclude $n=1$ or treat $f(1)=0$. ---