Problem
NT-B1-M04-P002 Inverse of 3
#2
★☆☆☆☆ Level 1 of 5
Solve \(3x\equiv1\pmod7\).
Find a number whose product with \(3\) is \(1\) modulo \(7\).
\(3\cdot5=15\equiv1\pmod7\). Hence \(x\equiv5\pmod7\).
First example of a modular inverse.