Solution

Part 1. These are the units digits of powers of 2:

1, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4

The pattern is that after the first digit, the digits repeat in cycles of 4. This statement is equivalent to

2ⁿ⁺⁴ - 2ⁿ = 0 (mod 10) for n > 1

(which is mathematical notation for 2ⁿ⁺⁴ - 2ⁿ is divisible by 10). This can be proved by factoring the left-hand side:

2ⁿ⁺⁴ - 2ⁿ = 2ⁿ(2⁴ - 1) = 15(2ⁿ)

15 is divisible by 5, and 2ⁿ is divisible by 2 (if n > 1), thus their product is divisible by 10.

Using Euler's Theorem (see the problem Powers Ending in 01) it can be shown that for all numbers that are relatively prime with 10, the units digits repeat in cycles of 4.

Part 2. The units digits of powers of 12 are the same as those of powers of 2, or:

1, 2, 4, 8, 6, 2, 4, 8, 6, 2, 4

Note that 12 = 2 + 10. Thus, by the Binomial Theorem,

12ⁿ = 2ⁿ + n(10)2^{n -1} + ...

All terms except the first are divisible by 10, thus the units digit is the same as that of 2ⁿ.

Part 3.

Note that

2002ⁿ = (2000+2)ⁿ = 2ⁿ + n(2000)2^{n - 1} + ...

has the same units digit as 2ⁿ. Further note that 2²⁰⁰³ has the same units digit as 2³, which is 8.

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