Question

斐波那契数列第2022项除以7的余数

Answer 1

斐波那契数列{Fn}中F1=F2=1,F<n+2>=Fn+F<n+1>,
以7为模，
n.......1....2.....3.....4....5....6.....7....8.....9....10....11...12....13....14...15....16....17...18
Fn≡...1.....1....2....3.....5....1.....6....0.....6....6......5......4......2.......6......1......0......1.....1,
可见{Fn}的以7为模的余数是以16为周期的，
2022=16*126+6，
所以F2022≡F6≡1,为所求。