Ta có 17 \(\equiv\) -1 ( mod 6 )
\(\Rightarrow\) 1720 \(\equiv\) (-1)20 \(\equiv\) 1 ( mod 6 )
Đặt 1720=6a+1 (a\(\in\) N)
Có: 146a+1 \(\equiv\) (146)ax14 ( mod 9 )
\(\equiv\) 1ax5 ( mod 9 )
\(\equiv\) 1x5 ( mod 9 )
\(\equiv\) 5 ( mod 9 )
\(\Rightarrow\) 14^17^20 chia 9 dư 5