Problem
NT-B2-M03-P011 Last Two Digits
#11
★★★★☆ Level 4 of 5
Find the last two digits of \(7^{100}\).
Check a short cycle modulo \(100\).
Compute \(7^2=49\), \(7^4\equiv49^2=2401\equiv1\pmod{100}\). Since \(100\) is divisible by \(4\), \(7^{100}=(7^4)^{25}\equiv1\pmod{100}\). The last two digits are \(01\).
Here it is better to find a short cycle than to apply \(\varphi(100)=40\).