what is 2 to the 10000th power as a number?

Well I would not try getting the actual value. 2^50 is already over 1 quadrillion. So, what I would do is this:

Look at first 5 powers of two.

They end in 2, 4, 8, 6, and then 2 again.

So, every fourth power ends in 2, starting at 2^1, so we take a multiple of 4 and add 1 to get the power of 2 that ends in two. 10000 is obviously divisible by 4, but since it is 1 less power than 10001, it ends in 6. There's one part completed. As for the number of digits, I am not sure how to get that. Sorry about that.

Here's how to determine the last digit:

The first 10 powers are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096...

If you look at the last digit of each number, you would see the repeating pattern: 2, 4, 8, 6, 2, 4, 8, 6...

If you divide 10,000 by 4 (the length of the repeating pattern), you get 2500. Since there are no remainders, this means the last digit falls on 6 (the end of the pattern). If you divided it and there was a remainder of 1, the last digit would be 2, a remainder of 2 means the last digit is 4 and so-on, and so-on.

Note the pattern of the last digits:

2¹ = 2

2² = 4

2³ = 8

2⁴ = 16

2⁵ = 32

2⁶ = 64

...

Last digit of 2ⁿ is

 2 if n ≣ 1 mod 4

 4 if n ≣ 2 mod 4

 8 if n ≣ 3 mod 4

 6 if n ≣ 0 mod 4

10,000 ≣ 0 mod 4, so last digit of 2¹⁰⁰⁰⁰ is 6.

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I found this formula but not its proof:

number of digits of 2ⁿ = n·log(2) + 1, rounded down to integer

log(2¹⁰⁰⁰⁰) = 10000log(2) = 10000×0.30103 = 3010.3

truncate to 3010

number of digits of 2¹⁰⁰⁰⁰ = 3010 + 1 = 3011

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I can explain the formula above.

Suppose a is the number of digits in x, and a > 1, then

 10ª⁻¹ ≤ x < 10ª

 10⁻¹ ≤ x/10ª < 1

Let x = k·10ª, then 10⁻¹ ≤ k < 10⁰

log(x) = log(k) + log(10ª) = log(k) + a

a = log(x) - log(k)

Since a is an integer and -1 ≤ log(k) < 0, a is the integer portion of log(x) plus 1.

I could not have explained the last digit any better than D.W. Yes, it is 6.

Now the number of digits:

2^10000

= 10^[Log(2^10000)]

= 10^[10000Log(2)]

3010 < 10000Log(2) < 3011

Therefore, 2^10000 has 3011 digits.

Deez nuts