A subset of the integers 1, 2, 3, 4, ……… 98, 99, 100 has the property that none of its members is 3 times another. What is the largest number of members such a set can have.
Answers
Answer:
The universal set is A containing the first 100 positive integers.
Set B is a subset of A. Set B has the property that if m and n are elements of B then the sum of m and n is not 125.
One possibility is that B contains the first 62 positive integers. Then the greatest number that can be formed by the sum of any two elements of B is 123.
The number of elements of set B is 62.
We can also selectively interchange one or more elements of B and and add that many selected elements of the complement of B.
e.g. We can remove 62 from B and add 63
or remove 60 and 61 from B and add 64 and 65
or remove 30, 35 and 40 from B and add 85, 90 and 95, and so on.
In all these cases the number of elements in set B would be 62.
Answer:
We have a set which has 20 elements (eg. {1,2,3,...,20}). We want to choose the maximum number of non-empty subsets we can such that none of them has more than 2 common elements. for example {1,2,3,4,5},{1,2,3,6,7,8} Has three common elements. For better clarification, if we wanted to solve this problem for 4 instead of 20 then the best way is to choose all the subsets except the {1,2,3,4} so the answer in this way is 15−1=14.