Problem
NT-B1-M07-P002 Last Digit of \(3^{25}\)
#2
★☆☆☆☆ Level 1 of 5
Find the last digit of \(3^{25}\).
The last digits of powers of \(3\) have period \(4\).
The cycle of last digits is \(3,9,7,1\). Since \(25\equiv1\pmod4\), the last digit is \(3\).
Make sure the student does not compute the huge power.