Question

Find the last digit of
(i) $(3{)}^{57}$ (ii) $(13{)}^{59}$

Answer 1

(i) The cyclicity of 3 is 4 Hence $\frac{57}{4}$ gives
the remainder 1 So the last digit of $3^{57}$ is same as
the last digit of $3^1$ i.e. 3
(ii) The number of
digits in the base will not make a difference to the last digit It is
last digit of the base which decides the last digit of the number itselt
For ${13}^{59}$ we find $\frac{59}{4}$ which gives a
remainder 3 So the last digit of ${13}^{59}$ is same as
the last digit of $3^3$ i.e 7