Python2, 188 187 bytes

def g(L):
 M=max(L);i=r=0;s=[0]*M;l=[1]+s[1:]
 while 1:
    if any(all((l+l)[o:o+min(L)])for o in range(M)):return~-s[r]*M+r
    if any(l[(i-a)%M]for a in L):l[i]=1
    else:r=i
    s[i]+=1;i=(i+1)%M

The second indentation is rendered as 4 spaces on SO, those should be tabs.

Actually a 'fast' solution, not bruteforce, uses 'Wilf's Method' as described here.