Demonic Gerbil

http://projecteuler.net/index.php?section=problems&id=5

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest number that is evenly divisible by all of the numbers from 1 to 20?


I actually solved this by hand because when I sat down with it today, I noticed how the number 2520 was constructed, and I attempted to repeat that with the larger number set. My reasoning is behind the cut, along with the answer, because by explaining my reasoning I make the answer plain as day anyway.

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I would like to revisit this with a program that does factorization for me, but I'm lousy about writing that sort of thing in an elegant manner so I probably will just leave this one solved with mathematical insight.

1+2=3
1+4+5+6+7=23

Which leads me to wonder if there is some more generalized rule for constructing these sorts of equations.

Take a set of digits, starting with 1, and going to n (such that n is an integer greater than 1). Can we always construct a summation such that the sum on the left side, using digits from that set, is equal to some combination (either as a sum, or as a two-digit number) of the remaining digits of the set?

I know the answer is 'no' for the pathological n=2 case, but I wonder, without sitting down and manually constructing an example for each other case, if there's a way to see if it's true or not.

(Further, I have a strong suspicion that the 1+4+5+6+7=23 construction, where I create a two digit number out of the unused digits, is unique.)

http://community.livejournal.com/mathematics/918429.html

x^b=b^x

I don't think I know enough Analysis to deal with the case where b is any real number. But for b is a positive integer, I worked out this approach:

Take the derivitive on either side with respect to x.

b*x^(b-1)=ln(b)*b^x

Repeat until the exponent on the left side is 0.

b(b-1)...(2)(1)=[ln(b)]^b*b^x

Of course that left side is just a factorial:

b!=[ln(b)]^b*b^x

Now some algebra gives us,

b!/[ln(b)]^b=b^x

ln{b!/[ln(b)]^b}=x*ln(b)

x=(ln{b!/[ln(b)]^b})/ln(b)

So, for b is a member of the natural numbers, we have solved the problem.

First: Good news. Got my QFT midterm back. 15/20 out on it. The average (of 7 students) was 12.6. I had the highest grade.

Second: Been skimming the titles and abstracts of papers from the LJ syndicated feeds I mentioned earlier. There's some interesting things in them from the archive.

Third: The latest Not Even Wrong is a pretty good read. But even better is his link to this paper by Bert Schroer which goes in depth into what he sees as errors in the last 10-20 years of 'modern' physics - and he doesn't cut people slack just because of known personality issues. Reminded me of reading about IRC in spots... I'm only part-way through that pre-print, but it's been an excellent read so far. You can hook up with the Not Even Wrong syndicated LJ feed feed if you want a dose of science in your friends list.

Fourth: Posted a comment over on the Mathematics list for an interesting problem. Hope I helped and didn't lead the guy astray.