What is more, since 74 has 01 as the final digit pair, the next four powers will end in 07,49,43, and then 01 again. Hence, as we compute succeeding powers, the pattern of the last two digits will simply repeat this cycle of length four, over and over again. To return to the question in hand, since 39=4×9+3, we will pass through this four-cycle nine times and then take three more steps in calculating the final two digits of 739, which must therefore be 43.And this works quite generally. In order tond the remainder when some power ab is divided by n say, we need only take the remainder r when a is divided by n and keep track of the remainders as we take successive powers of r. When we work with the remainder r, which will be a number in the range from 0 to n1, mathematicians say that we are working modulo n,discarding any higher multiples of n that may arise, as they leave a remainder of 0 when divided by n, and so cannot contribute to the value of thenal remainder r.