20051229, 21:32  #1 
Jun 2003
11000101011_{2} Posts 
Finding smooth numbers
Given a number n, how does one find the closest number that has all factors under a 1000? I can go and test each number, but is there a faster method?
What would be the running time of this faster algorithm? Is it in polynomial time? Thanks, Citrix 
20051229, 21:59  #3 
Jan 2005
Transdniestr
503 Posts 
I don't think you need to sieve even.
Define P as the lowest number that contains all the factors under some number N Define X as the first number greater than your input number M that also contains all the factors under N. then X= (1+int(M/P))*P ("int" rounds down) Finding the closest to M (either below or above) would just require a couple more simple steps. It doesn't really matter what the factors of P are either. (i.e. prime factors from 2 to N, all integers from 2 to N). The same idea would hold. Last fiddled with by grandpascorpion on 20051229 at 22:03 
20051229, 22:21  #4 
Jun 2005
Near Beetlegeuse
2^{2}·97 Posts 
So now you have a single equation in two unknowns. You can define X in terms of M and P, you can define M in terms of X and P or you can define P in terms of X and M. What now?
This also assumes that X and P exist as separate numbers. From Citrix question it is not clear (to me at least) whether he meant all of its factors < 1000, or all of the factors < 1000. If he meant the latter then P might be > M, in which case X might = P. Nice idea though. 
20051229, 22:32  #5 
Jun 2003
1579_{10} Posts 
What I meant was this consider n=100001 then the closest numbers with all factors below 1000 is 100000. Is there a method faster than sieving to find such numbers for any n?
Citrix 
20051229, 22:39  #6 
"Nancy"
Aug 2002
Alexandria
9A3_{16} Posts 
If you really want the bsmooth number closest to n, I think you'll need to sieve. If you want a bsmooth number "somewhere nearby" n, you can cut a few corners, i.e. by only looking for smooth multiples of a smooth fixed value, similar to what grandpascorpion suggested.
Alex 
20051230, 14:38  #7 
Jan 2005
Transdniestr
503 Posts 
Oops, I misinterpreted your question earlier, Citrix.

20051231, 09:37  #8 
Jun 2005
Near Beetlegeuse
388_{10} Posts 
No grandpascorpion, it is me that should apologise.
As Alex confirmed, you had the right idea. Whereas I was looking at your answer from my typically narrowminded algebraic perspective where a single equation in two unknowns has no solutions. I really should remember to shut my mouth when I don't know what I'm talking about. Sorry. 
20051231, 11:02  #9 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Finding smooth numbers.
Numbers: I really should remember to shut my mouth when I don't know what I'm talking about. Sorry./Unquote. Ha! ha! ha! ROTFL. You have finally admitted it old boy and that includes English also. Please, as a New year resolution remember your own dictum in the other posts as well!! Mally 
20051231, 11:07  #10 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Looks like Mally is completely losing it...
Alex 
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