forum.virtualchallengemeets.com

I recently came across this problem and I was wondering what methods you guys would use to solve it. I can show what I did later, but I'm hoping one of you finds a method I like better so I'm going to leave the answer open right now. It's a pretty simple problem but it was on a site that said "olympiad level problems so... who knows..."


How many digits are there in [unparseable or potentially dangerous latex formula]?

Maybe we could generalize this and everyone could share their methods for finding the number of digits in [unparseable or potentially dangerous latex formula].

Yeah I used the boring logarithm stuff. Then I decided I'd write a Java program with a while statement that kept dividing by 10 while the number was greater than or equal to 1.... basically doing a bunch of brute force work for me. Anyway, I'm lacking in creativity so I was hoping to hear of some other ways to solve that.

Well, 2^10's about 1000, which means we get 2^1000 = (1000)^100 = 10^300. So I'd say 301 digits plus a couple from yon approximation; whats an order of magnitude or two between friends anyway?

this gets it down to two possibilities

[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]

showing that [unparseable or potentially dangerous latex formula] isn't too bad, but ummm i think to win you'd have to get [unparseable or potentially dangerous latex formula] which is a little harder/actually still leaves you with 2 cases =(

cout<<pow(2,1000);
Count.

this question doesn't really need that many process,
only 2 steps will do it:
1000log2=301.0299957
so [unparseable or potentially dangerous latex formula]
so it has 302 digits.
I don't think it's any olympiad kind of question...