Remainders Revisited #4

Number Theory Level pending

$\Large 4^{47^ {23}} \bmod 25 = \, ?$

8 14 3 24

1 solution

Maggie Miller
Aug 2, 2015

Note $4^5=1024\equiv -1\pmod{25}$ . Moreover, mod 10

$47^{23}\equiv 7^{23}\equiv (-1)^{11}\cdot7\equiv-7\equiv 3\pmod{10}$ .

Therefore, there exists an integer $n$ so that $47^{23}=10n+3$ . Then mod 25

$4^{47^{23}}=4^{10n+3}=(4^5)^{2n}\cdot64\equiv (-1)^{2n}\cdot 14\equiv\boxed{14}\pmod{25}$ .