What integer \(n\) satisfies \(0\le n<{101}\) and \(100n\equiv 72\pmod {101}~?\)
Using the Euclidean Algorithm, I found n = 31.
101 - 72 = 29.
\(\displaystyle 29 \equiv -72 \bmod101,\\ 100 \equiv -1 \bmod 101, \\ \text{so}\\ 2900 \equiv 72 \bmod 101.\)
+216 
