Problem
NT-B1-M07-P003 A Power of Four
#3
★☆☆☆☆ Level 1 of 5
Find the remainder of \(4^{12}\) modulo \(7\).
Check what \(4^3\) is modulo \(7\).
\(4^2=16\equiv2\pmod7\), \(4^3\equiv8\equiv1\pmod7\). Hence \(4^{12}=(4^3)^4\equiv1\pmod7\).
The short period \(3\) is better than a general theorem here.