The last digit !!!!!!!!

Look this pattern:

${2003}^{1}$ unit digit is 3

${2003}^{2}$ unit digit is 9

${2003}^{3}$ unit digit is 7

${2003}^{4}$ unit digit is 1

After 4 powers the sequence repeats. ${2003}^{5}$ unit digit is 3

${2003}^{6}$ unit digit is 9

...

The remainder left when 2002 is divided by 4 is 2. ${2003}^{2002}$ the unit digit is 9 according to the pattern.

The unit digit of 2001 is 1. 1 times 1 is 1. Logically ${2001}^{2002}$ the unit digit is 1.

$9 + 1 = 10$ the unit digit of ${2003}^{2002} + {2001}^{2002}$ is 0.

2003^2002 + 2001^2002≡? mod 10 I) (2003,10) = 1 => φ(10)=4 2003^4≡1 mod 10 => 2003^2000≡1 mod 10=> 2003^2002≡9 mod 10

II) (2001,10)=1=> φ(10)=4 2001^4≡1 mod 10 => 2001^2000≡1 mod 10=> 2001^2002≡1 mod 10

∴ 2003^2002 + 2001^2002≡ 9 +1=10 ≡ 0 mod 10

Logically in 2003 the last digit is 3 Powers of 3 gives many numbers(as last digit) let's make it 9 As it is last digit don't matter about others so 3 raised to 2002 =9 raised to 1001 In odd Powers of 9 last digit is 9 and even Powers gives 1. And in 2001 last digit is 1 so there any power will have last digit 1 So 9+1=10 so last digit =0

$2003^{2002}=10 B + 3^{2000} \times 9 =10 B + (80+1)^{500} \times 9 = 10 A + 9$ $2001^{2002}=(2000+1)^{2002}=10C+1$ $\Rightarrow 3^{2002} + 2001^{2002} = 10 D \Rightarrow R=\boxed{0}$