The Magic of extraordinarily huge numbers

Find the last digit. 2012^2009 ==> 2^9 = 512, last digit of this is 2. 2009^2012==> 9^2 = 81, last digit of this is 1. 2012^2009 + 2009^2012 | 2 + 1 = 3

Number 2 and 9 are coming from the last digits of 2012 and 2009. To find the last digit number, we must use all the last digit.

Since we're considering the last digit, we should apply $A^B \bmod {10} = A^{B \bmod 4} \bmod{10}$ . Consider modulo $10$ :

$2012^{2009} \equiv 2^{2009} \equiv 2^{2009 \bmod 4} = 2^1 = 2$

$2009^{2012} \equiv (-1)^{2012} \equiv (-1)^\text{even number} = 1$

Add them up: $2 + 1 = \boxed{ 3}$

P(10) = 10(1 - 1/2)(1 - 1/5) = 4

2012^2009 = (10k + 2)^(4k + 1) = 2^1= 2

2009^2012 = (10k +9)^(4k ) = 9^0 = 1

The remainder when dividing by 10 = 2 + 1 = 3

Then the last digit = 3

to find the last digit divide the power by 4 then reminders can be 1,2,3 or 0 then if the reminder is 1 then last digit =x^1 if the reminder is 2 then last digit=x^2 similar for 3 but when reminder is 0 ,last digit=x^4 now, in question 2012^2009 by applying our trick we have to divide 2009 by 4 so reminder=1.therefore,last digit=2^1 that is=2 2009^2012, for this we have to divide 2012 by 4 so reminder will be 0.therefore last digit=9^4 that is =1. by adding 1 and 3 we got answer=3