power of 2

We have $2^{4} \equiv 6 (mod 10) \Rightarrow 2^{2007} = [(2^{4})^{501}] \times 2^{3} \equiv 6 \times 8 \equiv 8 (mod 10).$

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A cycle works for exponents of 2, it comes like 2, 4, 8, 6, 2, 4, 8, 6 ... So, after 2^2004 it gives 6 as the last digit, and for 2^2007 it will must give 8 !

2^2007 can be written as 2^4(501)+3. Since we know that the last digit in the expansion of 2^4n+3 is 8. Therefore the last digit in the expansion of 2^2007 is 8.

In the power of two 2, the last digit is repeated 2 ; 4 ; 8; 6 .

2007 mod 4 (four terms: 2 4 8 6) = 3

2 is the first term, 4 is the second term, 8 is the third term.

so the solution is 8.