Using the 7-adic Euclidean Algorithm for $\frac{181625}{11}$ and $\frac{9360}{11}$:
- $\frac{181625}{11} = (2)\left(\frac{10555}{2}\right) + \left(\frac{9360}{11} \cdot 7^1\right)$
- $\frac{10555}{2} = \left(\frac{12}{7}\right)\left(\frac{9360}{11} \cdot 7^1\right) + \left(\frac{-2215}{22} \cdot 7^2\right)$
- $\frac{9360}{11} \cdot 7^1 = \left(\frac{-10}{7}\right)\left(\frac{-2215}{22} \cdot 7^2\right) + \left(\frac{-5}{11} \cdot 7^4\right)$
- $\frac{-2215}{22} \cdot 7^2 = \left(\frac{50}{49}\right)\left(\frac{-5}{11} \cdot 7^4\right) + \left(\frac{-5}{22} \cdot 7^5\right)$
- $\frac{-5}{11} \cdot 7^4 = \left(\frac{2}{7}\right)\left(\frac{-5}{22} \cdot 7^5\right) + 0$
If someone could guide me to understand just one term, like $\frac{12}{7}$, I would greatly appreciate it!
Applying the 5-adic Euclidean Algorithm for 325 and 35:
- $325 = (0)(35) + (13)(5^2)$
- $35 = \left(\frac{-11}{5}\right)(13)(5^2) + (6)(5^3)$
- $(13)(5^2) = \left(\frac{-2}{5}\right)(6)(5^3) + (1)(5^4)$
- $(6)(5^3) = \left(\frac{6}{5}\right)(5^4) + 0$
If someone could explain the derivation of any single term, e.g., $\frac{-11}{5}$, I'd be very grateful! Thank you.