Mind Calculator..!!

Fermat's Little Theorem states that:

$a^{p-1} = 1$ (mod p)

For co-prime a and p. So:

$7^{9} = 1$ (mod 10)

Thus:

$7^{39} =7^{3} = 3$ (mod 10)

9 multiplied by 7 gives 63, so the last digit of the multiplication in question is obviously 3.

since 39 is divisible by 3, then in implies that 727^3 is the same with the last digit of 727^39 which is 3.................Using To-ongs cyclic theory

last digit will be the last digit of 7*9 (=63) that is 3 the digits in the ten's place in the sum hav no effect on the unit's place of the answer. even if you multiply by the normal method you will find that he unit's place is not affected by the numbers at the ten's place in the question

Last digit of 727 is 7. If we multiply 7 7 7 7 the last digit comes out to be 1. Further if we multiply 7 again we will again end up with a number whose last digit is 1. Hence taking 4 group 7s we at last get 1 till 727^36. Hence the final answer will be (considering the last digits) = 1 7 7 7 = 3(Last digit)