I swear this is the last

Author: brebenelmihnea

content/4_Gold/Modular.mdx

@@ -292,7 +292,7 @@ assert 2 * x % MOD == 1
 We can also find modular inverses through Euclidean Division.
 Given the prime modulus $m > a$ we have:
 $m = k \cdot a + r$, where k = $floor(\frac{m}{a})$	and $r = m \mod a$. Then: $0 = k \cdot a + r \mod m \iff r = -k \cdot a \mod m
-\iff r \cdot a^{-1} = -k \mod m \iff a^{-1} = -k \mod m \cdot r^{-1} \mod m$.
+\iff r \cdot a^{-1} = -k \mod m \iff a^{-1} = -k \cdot r^{-1} \mod m$.
 
 Here is a short recursive implementation of the above formula: