Problem
NT-B1-M10-P016 Sum of Powers
#16
★★★☆☆ Level 3 of 5
Find the last two digits of \(3^{2026}+7^{2026}\).
The powers of \(3\) and \(7\) modulo \(100\) have period \(20\).
Since \(3^{20}\equiv1\pmod{100}\) and \(7^{20}\equiv1\pmod{100}\), and \(2026\equiv6\pmod{20}\), we get \(3^{2026}\equiv3^6\equiv29\) and \(7^{2026}\equiv7^6\equiv49\pmod{100}\). Their sum is \(78\).
CRT is possible, but the direct cycle is shorter here.