2018 is special
is a subset of { }. Every pair of elements in are relatively prime to one another. If , (i.e., no. of elements), then find the least number of primes in .
The answer is 16.
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1 solution
Notice 2 0 1 8 <45 and there are 14 prime no.s from 1 to 44.Any composite number N must have a prime factor from 2 to 44.So any composite no. of S must have a prime factor within these 14 primes.Also all members are coprime to each other.Now for making all coprime pairs there can be at max 14 composite no.s and hence at least 16 prime nos.