1. **Stating the problem:** 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 $m$, i.e., the highest power of 3 dividing $m$. We want to find $$ S = \sum_{n \in A} f(n). $$ 2. **Understanding the problem and formula:** - The 3-adic valuation $\nu_3(m)$ counts the exponent of 3 in the prime factorization of $m$. - The set $A$ consists of integers from 1 to 243 that are coprime to 3, so $n$ is not divisible by 3. - We want to sum $\nu_3(n^{81} - 1)$ over all such $n$. 3. **Key observations:** - Since $n$ is coprime to 3, $n$ is invertible modulo powers of 3. - The number 243 is $3^5$, so $1 \le n \le 3^5$. - The function $f(n) = \nu_3(n^{81} - 1)$. 4. **Using Lifting The Exponent (LTE) lemma:** For odd prime $p=3$, and integers $a,b$ with $p \mid (a-b)$ and $p \nmid ab$, and $k$ odd, $$ \nu_3(a^k - b^k) = \nu_3(a-b) + \nu_3(k). $$ Here, set $a=n$, $b=1$, and $k=81$ (which is $3^4$). Since $n$ is coprime to 3, $3 \nmid n$, and $3 \mid (n-1)$ if and only if $n \equiv 1 \pmod{3}$. 5. **Evaluate $\nu_3(n^{81} - 1)$ casewise:** - If $n \equiv 1 \pmod{3}$, then $3 \mid (n-1)$, so LTE applies: $$ f(n) = \nu_3(n-1) + \nu_3(81) = \nu_3(n-1) + 4. $$ - If $n \not\equiv 1 \pmod{3}$, then $3 \nmid (n-1)$, so $\nu_3(n^{81} - 1) = 0$. 6. **Rewrite the sum $S$:** Only $n$ with $n \equiv 1 \pmod{3}$ contribute. The set $A$ has all $n$ with $1 \le n \le 243$ and $3 \nmid n$. Among these, those with $n \equiv 1 \pmod{3}$ are exactly those with $n \equiv 1 \pmod{3}$ and $3 \nmid n$ (which is automatic). Number of such $n$ is the count of numbers from 1 to 243 congruent to 1 mod 3. These are $n = 1,4,7,\dots, 241$. Count is $\frac{243 - 1}{3} + 1 = 81$. 7. **Express $S$ as:** $$ S = \sum_{\substack{1 \le n \le 243 \\ n \equiv 1 \pmod{3}}} \left( \nu_3(n-1) + 4 \right) = \sum_{\substack{1 \le n \le 243 \\ n \equiv 1 \pmod{3}}} \nu_3(n-1) + 4 \times 81. $$ 8. **Change variable:** Let $n = 3k + 1$ with $k=0,1,\dots,80$. Then $$ \nu_3(n-1) = \nu_3(3k) = 1 + \nu_3(k) \quad \text{for } k \ge 1, $$ and for $k=0$, $n=1$, so $n-1=0$, and $\nu_3(0) = \infty$, but since $n=1$ is in $A$, we consider $f(1) = \nu_3(1^{81} - 1) = \nu_3(0) = \infty$. But $f(1)$ is infinite, so the problem likely excludes $n=1$ or interprets $\nu_3(0)$ as infinite. Since $n=1$ is in $A$, and $f(1)$ is infinite, the sum $S$ would be infinite. Assuming the problem means $n \neq 1$ or $\nu_3(0)$ is not counted, we exclude $n=1$. So sum over $k=1$ to $80$. 9. **Sum becomes:** $$ S = \sum_{k=1}^{80} \left(1 + \nu_3(k) + 4 \right) = \sum_{k=1}^{80} (5 + \nu_3(k)) = 5 \times 80 + \sum_{k=1}^{80} \nu_3(k) = 400 + \sum_{k=1}^{80} \nu_3(k). $$ 10. **Calculate $\sum_{k=1}^{80} \nu_3(k)$:** This is the total exponent of 3 in the factorial $80!$. Using Legendre's formula: $$ \sum_{k=1}^{80} \nu_3(k) = \nu_3(80!) = \left\lfloor \frac{80}{3} \right\rfloor + \left\lfloor \frac{80}{9} \right\rfloor + \left\lfloor \frac{80}{27} \right\rfloor + \left\lfloor \frac{80}{81} \right\rfloor = 26 + 8 + 2 + 0 = 36. $$ 11. **Final value:** $$ S = 400 + 36 = 436. $$ **Answer:** $$ \boxed{436} $$