Determine the last two digits of 9889SolutionSince we are on

Determine the last two digits of 98^89.

Solution

Since we are only interested in last two digits so we take larger and larger powers of 98 and at each step ignoring all digits other than the last two digits.

We perform repeated squaring and at each step ignoring the digits other than last two digits.

98^2=9604=04(ignoring digits other than last two)

98^4=04^2=16

98^8=16^2=256=56

98^16=56^2=3136=36

98^32=36^2=1296=96

98^64=96^2=9216=16

98^80=98^(64)*98^(16)=16*36=576=76

98*88=98^(80)*98^8=76*56=4256=56

98*89=98^(88)*98=56*98=5488=88

Hence last two digits of:98^89 are 88