My problem is the following, for an application i need the lowest number which can be divided bij the following set of numbers: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15.
My first sollution was 15! - 7! since all number which are undernearth 8 (so 7 and lower) can allways be divided by when dividing by a higher number. I know i can also divide this outcome by 10 and make it still dividable, but i think there shouldeven be a better way to do this.
For each prime, you just need to find the maximum number of factors that are in one number. For $2$, that is $3$, coming from $8$, for $3$ you need $2$ because of $9$. All other primes come once each. If you need it for other top values, they are given in OEIS A003418
This should work in general. Let $p$ denote a prime.
$$LCM(1,...,N) = \prod_{p\le N} p^{\lfloor\log_{p}(N)\rfloor} $$
For $N=15$ we have
$$LCM(1,...,15) = \prod_{p\le 15} p^{\lfloor\log_{p}(15)\rfloor} = (2^3)(3^2)(5)(7)(11)(13)$$
