What is the
last digit of 32^{91}?

(a) 2 (b) 4 (c) 6 (d) 8

To obtain
the last digit of 32², 32³, ..., multiply the last digits i.e., 2x2, 4×2, 8x2
etc.

For 32^{1},Last digit = 2, i.e., 2

For 32^{2}, Last digit = 2 x 2 = 4, i.e., 4

Hence, the last digit is given by:

For 32³,
Last digit = 4 x 2 = 8, i.e., 8

For 32^{4},
Last digit = 8 x 2 = 16, i.e., 6

For 32^{5},
Last digit = 6 x 2 = 12, i.e., 2

For 32^{6},
Last digit = 2 x 2 = 4, i.e., 4

For 32^{7},
Last digit = 4 x 2 = 8, i.e., 8

For 32^{8},
Last digit = 8 x 2 = 16, i.e., 6

Every 4th

We can
observe that the unit digit gets repeated after power of 32.

Hence, we
can say that 32 has a power cycle of 2, 4, 6, 8, with frequency 4.Therefore,
for powers, 1, 5, 9, 13,...., the last digit will be 2. For powers, 2, 6, 10,
14,...., the last digit will be 4.

For powers,
3, 7, 11, 15,...., the last digit will be 8.

For powers,
4, 8, 12, 16,...., the last digit will be 6.

In the
question, the power is 91.

91 is a number that gives a remainder 3 when divided by 4.

Therefore, 91 is in the series 3, 7, 11, 15...

The last digit is 8. Hence option (d).