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You may want to check here: http://www.ticalc.org/pub/83plus/basic/math/algebra/

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It is easy to calculate that 2^10 = 1024 (at least for me, I’m a computer guy and many computer guy memorizes the series of number 2^n since base-2 is very important for us)

Still that isn’t so helpful. 2^1024+1 is a very large number.

So how do we find it?
Ask the question, do we need to know the number to find its number of digits?

If you answer „NO“ with confidence, you’re in the right track.

So, how do we find out the number of digits a number occupies without knowing the number itself?

Turns out there is a trick using logarithm:

log(n) / log(10) will tell you the number of digits of a number in base 10.
Let’s see: log(1024) / log(10) = 3.0102…
Rounding that UP, we get 4. (note: you *always* round up, even if it is exactly 3; an alternative way to say this is, you add 1 and round down)

Let’s try again:
log(23443245) / log(10) = 7.37…
Rounding that up, we get 8. Last time I checked 23443245 has 8 digits.

So, why does it works? Remember the definition of logarithms:
x = log(n) / log(10)
is just another way to say:
x = log base 10 of n
i.e.
10^x = n

Since 10^1 = 10, 10^2 = 100, 10^3 = 1000, etc… Then log(n) / log(10) can give us the number of digits.

Now back to the problem.

The number 2^1024+1 is difficult to calculate, and we do not want to have to calculate it.

However, notice that in base-2 (binary), the number 2^1+1 = 11[2], 2^2 = 101[2], 2^3 = 1001[2], … 2^(10) = 10000000001[2], etc…

I.e. The number of digits of 2^n in base-2 is n rounded UP.

Now, it happens that our trick of counting digits with logarithms also works with binary, with a liiiitle modification:
y = log(n) / log(2)

Previously we have shown that 2^1024+1 will have 1025 digits when written in base-2.
Therefore, we know that when n = 2^1024+1, y = 1024.xxx (1024 rounded up is 1025, since 2^1024 and 2^1024+1 is very near and later we see that it is far from the base-10 number-of-digit boundary, it is easy to see that using y =1024 won’t change the number of digits)
1024 = log(n) / log(2)
log(n) = 1024 * log(2)

We also already established that x = log(n) / log(10) will gives us the number of digit of n in base 10
x = log(n) / log(10)
log(n) = x * log(10)

Combining those two, we get
1024 * log(2) = x * log(10)
so solving that
x = 1024 * log(2) / log(10)
x = 1024 * log(2) / log(10)
x = 308.254…
Rounding up, we know that 2^1024 will have 309 digits when written in base 10.

2. I don’t understand question 2. It is always possible to bisect an arbitrary angle with a straight edge and a compass; therefore the number of degrees can always be halved even though it may be impractical at some stage. Maybe I’m misunderstanding the question.

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LOL funny. I went to computer science course open day in University, and i was amazed ! None of the guys were ugly or anything. Everyone looked normal , and quite a few were hot ! 🙂 and i was the only girl in that open day . It was well fun . I wanna do that course

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I got:
weird books
pencil case
lunch bag
coloured pencils
pastels
nail kit thing (kinda kool)
camera (awesome)

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55 reads the same when you turn the calculator upside down.

66 reads as a number, all others read as letters when you turn the calculator upside down.