Problem
NT-B1-M07-P013 Sum of Two Large Powers
#13
★★★☆☆ Level 3 of 5
Find the remainder of \(2^{2026}+3^{2026}\) modulo \(5\).
Both powers have period \(4\) modulo \(5\).
Since \(2026\equiv2\pmod4\), \(2^{2026}\equiv2^2\equiv4\pmod5\) and \(3^{2026}\equiv3^2\equiv9\equiv4\pmod5\). The sum is \(4+4=8\equiv3\pmod5\).
Here each base must be handled separately.