Question

Ones digit of 2 to the power of 2029?

Answer 1

This is given by the value of

`2^(2029) \pmod{10}`

which is an integer
`n` between 0 and 9 such that

`2^(2029) = 10m + n`

where
`m` is any integer.

Any power of 2 ends in an even digit, so
`n` must even. Write
`n=2n'` for integer
`n'`. Then

`2^(2029) = 10m + 2n' \implies 2^(2028) = 5m + n'`

so now we want to find

`2^(2028) \pmod 5`

We have
`\phi(5) = 3`, where
`\phi(n)` is Euler's totient function, the number of primes less than or equal to
`n`. Then

`2^(2028) \equiv 2^(3\cdot676) \equiv \left(2^(676)\right)^3 \equiv 1 \pmod 5`

by Euler's theorem, and

`n'=1\implies n=2 \implies 2^(2029) = 10m + 2 \iff 2^(2029) \equiv \boxed{2} \pmod{10}`

Answer 2

Answer: 8

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Step-by-step explanation:

Let's list out a few numbers of the form 2^n where n is some positive integer.

• 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

Then focus solely on the units digit of each result:

2, 4, 8, 6, 2, 4, 8, 6, 2, ...

We see the pattern is "2,4,8,6" and that pattern repeats forever. The reason why is because once we reach 6, we get back to 2 and it restarts the cycle. The pattern repeats itself every 4 values.

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So we'll divide the exponent by 4 and look at the remainder (see note at the end)

2019/4 = 504 remainder 3

The remainder 3 tells us to look at the 3rd slot of the list {2,4,8,6}

That third item is 8

Therefore units digit of 2^2019 is 8.

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Side note: if you get a remainder of 0, then look at slot 4. This works because 4/4 = 1 remainder 0.

In other words, think of "remainder 0" as "remainder 4".