Q: Let n = 2^2008 + 2008^2. What is the units (ones) digit of 2^n + n^2?

A: This is related to the powers of two and also the last digits of squared numbers.

We first have to find n

Since the last digit of 2^(a power) has a pattern we can figure this out!

2^2008

2 to a power has a last digit pattern

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

2^7 = 128

2^8 = 256

2^9 = 512

2^10 = 1024

2^11 = 2048

2^21 = 2097152

2^31 =

Since it is a cycle of 4, every power divisible by 4 will end with a 6…therefore 2^2008 ends with a 6…but we actually need to know the **last two digits** as 2^1 and 2^11 do not share the same last digit but 2^1 and 2^21 do. So when we think about very large exponents for powers of 2, we need to know the last two digits…

2^4 ends with 16

2^8 ends with 56

2^12 ends with 96

2^16 ends with 36

2^20 ends with 76

2^24 ends with 16

2^28 ends with 56 and so on

every multiple of 20 ends with 76

so 2^2000 would end with 76

2^2004 ends with 16

2^2008 ends with **56**

2008^2 =40320**64**

Adding up 64 + 56 = 120 so the last two digits of n would be **20**

so in evaluating 2^n + n^2

First 2^n: 2^20 or 2^120 or 2^ 220 always ends in a “**6**”

then n^2, anything with a 0 in the ones digit squared would have a ones or units digit of zero.

Therefore, the last digit would be **6**.