Question

the last digit of 253^1002

Answer 1

$mod((253^{0}),10) = 1 \\ mod((253^{1}),10) = 3 \\ mod((253^{2}),10) = 9 \\ mod((253^{3}),10) = 7$

this pattern would get repeated you can check it for yourself for more powers
So generalized pattern is
$mod((253^{4n+0}),10) = 1 \\ mod((253^{4n+1}),10) = 3 \\ mod((253^{4n+2}),10) = 9 \\ mod((253^{4n+3}),10) = 7$

as 1002 = 4(250) + 2
$mod((253^{1002}),10) = mod((253^{4(250)+2}),10) = 9$

Answer 2

Answer:

multiply and get the answer

don't be lazy to look

brainly