Is the decimal representation really finite?

Finding the last non-zero digit is same as finding the last digit of $\frac { 1 }{ { 5 }^{ 2003 } } \times { 10 }^{ 2003 }$ . This can be simplified as

$\frac { { 10 }^{ 2003 } }{ { 5 }^{ 2003 } } ={ 2 }^{ 2003 }$

Last digit of $2^x$ follows a cyclic pattern of period 4. Hence the last digit of $2^{2003}$ is same as the last digit of $2^{3}$ , which is 8.

Notice that $\frac{1}{5^n}$ in decimal form is always: 0. | n-1 number of 0 | 2^n.

$\phi(10)=4$

$2^{2003}\equiv2^3 = 8 \pmod{10}$

The last digit of decimal representation of $5^{-n}$ follows the cyclic order 2,4,6,8,2... So for 2003, it is 8