As given above this problem can be solved using the look-for-pattern strategy, but the guess and check strategy can also be implemented to solve the same problem.

Suppose 𝑥 is the last digit of 2^2015, As 2^2015 is an even number, then 𝑥 =0,2,4,6,8. Consider the number “ 2^2015 – 𝑥 ”, which has 0 (zero) in its unit place. Alternatively, we will check here which 𝑥 value gives 10|( 2^2015 – 𝑥).

Next, we check whether the following numbers are divisible by 10.

2^2015−0, 2^2015−2, 2^2015−4, 2^2015−6, 2^2015−8.

Among these five numbers, only one number is divisible by 10. ( 2^2015 has only one unique digit in its unit place). All the above numbers are divisible by 2.

Next, check which numbers are divisible by 5.

we know that 𝑥+𝑎|𝑥^n+𝑎^n for any odd integer 𝑛. If we carefully observe and choose values for 𝑥=4,𝑎=1, and 𝑛=1007 (odd).

4+1|4^1007 +1

4+1|2^2014 +1
⇒ 5|2(2^2014 +1)
5|2^2015 +2.

We can use this to check which numbers going to be divisible by 5.

As we know 5 | 2^ 2015 +2, within every five units from that number, is divisible by 5. So here if we move 5 units back from 2^ 2015 +2 we come across 2^ 2015 −3. In the same way, if we go 5 units back we will end up in 2^ 2015 −8. Which means 5|2^ 2015 −8.

2 and 5 are relatively prime numbers, therefore 10|2^ 2015 −8.

Among these numbers 2^ 2015, 2^ 2015−2, 2^ 2015−4, 2^ 2015−6, 2^ 2015−8 only 2^ 2015 −8 is divisible by 10. therefore 8 is the last digit of 2^ 2015. QED