Question

find the last digit (32)^32

Answer 1

(32)³² = (2⁵)³² = 2¹⁶⁰

Therefore (32)³² can be rewritten as 2¹⁶⁰

The exponent of 2 has a regular pattern:

2¹ = 2                           ---------------------- Pattern [ 1 ]

2² = 4                           ---------------------- Pattern [ 2 ]

2³ = 8                           ---------------------- Pattern [ 3 ]

2⁴ = 16 (last digit = 6)  ---------------------- Pattern [ 4 ]

And then it repeats

Find the last digit of 32³²:

160 ÷ 4 = 40

⇒ It will fall into Pattern [ 4 ]

⇒ the last digit is 6

Answer: The last digit is 6

Answer 2

here is your answer OK

32 = 2^5
32^32 = (2^5)^32 = 2^(5x32) = 2^160
(32^32)^32 = (2^160)^32 = 2^(160x32) = 2^5120

We observe that
2^1 ends in 2
2^2 ends in 4
2^3 ends in 8
2^4 ends in 6

2^5 ends in 2
2^6 ends in 4
2^7 ends in 8
2^8 ends in 6

and so on.

We may conclude (for a non-negative integer k) that
2^(4k+1) ends in 2
2^(4k+2) ends in 4
2^(4k+3) ends in 8
2^(4k) ends in 6

5120 = 4(1280) = 4k, where k = 1280.

Thus, (32^32)^32 = 2^5120 ends in 6.