Recent content by nugae

I'm going to answer the question for 1000! / 5^n. You should be able to see that the answer is the same for 1000!/10^n. Numbers from 1 to 1000 that are divisible by 5 each contribute a factor of 5 to 1000!. That gives us 1000/5 factors of 5 so far. Numbers from 1 to 1000 that are divisible...

Project a pig P from a spacecraft in near-Earth orbit. Assume that P is rotating rapidly (or, equivalently, that it is perfectly conductive). What temperature will P reach when it is in radiative equilibrium with the Sun on the one hand and the CRB on the other? The Stefan-Boltzmann law...

Lyx is free and works well. I've used it to write the odd paper, and you can also cut and paste formulae from it into this forum.

This is why I hate posting answers in forums. I think of a nice abstruse proof, then do my best to put it simply (and long-windedly)... and then someone else comes up with a truly excellent, simple, intuitive solution that doesn't require any clever algebraic symbols at all! Thank you...

It's fairly easy to see that the problem says "the door is closed if the number has an odd number of factors, the door is open if it has an even number of factors" - where we define "the number of factors" of a number N as including all numbers that divide N, including 1 and N itself. For...

I've been looking at the value N(n) of N that satisfies the equation \sum_{1}^{n}(N-i)^{n}=N^{n} Thus turns out to be N(n)=1.5+\frac{n}{ln2}+O(1/n) where the O(1/n) term is about 1/400n for n>10. I've verified this by calculation up to about n=1000, using Lenstra's long...

Binary numbers are inconveniently long for human beings to handle (though not as bad as unary ones!) so grouping the digits makes life more comfortable. Visually it's easier to count fewer groups of 4 digits rather than 4 times as many individual digits. It's the same reason that in many parts...

A new question I'm a bit confused about when you post a question here and when you start a new thread, so if anyone wants me to make a new thread out of this, just tell me and I'll do it! I've been looking at the value N(n) of N that satisfies the equation \sum_{1}^{n}(N-i)^{n}=N^{n} Thus...