N = 1^{1986} + 2^{1986 }+ 3^{1986} + 4^{1986}.

What is the last digit of N?

**Answer**

To find the last digit on N we have to find the last digit of each of the terms adding up to N: 1^{1986}, 2^{1986}, 3^{1986}, and 4^{1986}. Of course, one could always compute the result one power at a time. Steady and slooowly. The world record for the fastest time to complete the official 250-piece jigsaw puzzle of the Guinness World Records is 13 minutes and seven seconds. Mathematics shouldn’t necessarily be a race against time, but it should be about finding the answer in the most efficient, i.e. interesting, way possible.

**Let’s start with 2 ^{1986}. **

Can we write it as product of factors whose last digits are easy to determine?

What number, power of 2, yields the same last digit when raised to any power?

It’s 16 = 2^{4}. 16 raised to any power will give a number that ends in 6 (because 6 raised to any power gives a number that ends is 6. Always.)

You can rearrange 2^{1986} as:

2^{1986} = 2^{4×496+2} = 2^{4×496} x 2^{2} = (2^{4})^{496} x 4 = 16^{496} x 4

16^{496} ends in 6, and 6 x 4 = 24, so the **last digit of 2 ^{1986} is 4.**

**How about 4 ^{1986}? **

It’s very similar to 2^{1986}. Try it on your own, and then check your answer below.

4^{1986} = 4^{2×993} = 16^{993}.

** Last digit of 4 ^{1986 }is 6.**

**How about 3 ^{1986}? **

Try it on your own, and then check your answer below.

3^{1986} = 3^{4×496+2} = (3^{4})^{496} x 3^{2} = 81^{496} x 9.

Since 81^{496} ends in 1, then **last digit of 3 ^{1986} is 1 x 9 = 9.**

Last digit of N is last digit of 1 + 4 + 9 + 6 = 20. It’s just … 0.