Problem
NT-B1-M07-P015 Last Two Digits of \(9^{2026}\)
#15
★★★☆☆ Level 3 of 5
Find the last two digits of \(9^{2026}\).
Compute the cycle of powers of \(9\) modulo \(100\).
\(9^1\equiv9\), \(9^2\equiv81\), \(9^3\equiv29\), \(9^4\equiv61\), \(9^5\equiv49\), \(9^6\equiv41\), \(9^7\equiv69\), \(9^8\equiv21\), \(9^9\equiv89\), \(9^{10}\equiv1\pmod{100}\). Since \(2026\equiv6\pmod{10}\), \(9^{2026}\equiv9^6\equiv41\pmod{100}\). The last two digits are \(41\).
The period is not always \(\varphi(100)\); sometimes it is shorter.