where the summation is over all subsets {i1, ... ,i\mu } of {1, ... ,t } and for the empty subset the product is considered to be one. The result is best possible, for every N there are f(N) divisors no two of which are relatively prime. Theorem 2. Let G1, ... ,Gm be m distinct divisors of N not two of which are relatively prime. Assume m < g(N). Then there are g(N)-m further divisors Gm+1, ... ,Gg(N) so that no two of the g(N) distinct divisors Gi, 1 \leq i \leq g(N) are relatively prime. Theorem 2 is best possible.
Classif.: * 05A17 Partitions of integres (combinatorics)
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