May 2019 Brain Puzzler Solution

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 =4032064
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.

 

 

 

 

 

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