20. Consider a bowl containing marbles labeled with natural numbers. Suppose that 30 marbles are labeled with numbers divisible by 6, 20 marbles are labeled with numbers divisible by 7, and 10 marbles are labeled with numbers divisible by 42. What is the smallest possible number of marbles in this bowl? So I have included this information in a Venn diagram where I have three sets in blue elements divisible by 7, we have 20 of them, in B elements divisible by 6 and there are 30 of them, and in red we have elements divisible by 42 and we have 10 of them. And the bowl would be the union of all these sets. We have to minimize the size of this union. Now note the set containment here. B and C intersect. Normally there are numbers both divisible by 7 and 6 and in that intersection I have placed the set A because anything that will be divisible by 42 will also be divisible by the factors of 42 which are 6 and 7. So we have that the set A belongs to the intersection of B and C. Now let an element belong to the intersection of B and C. Then we see that X is divisible by 6 and by 7. So X has to be divisible also by the least common multiple of 6 and 7 which would be 42. Then we see that X belonging in the intersection of B and C and being divisible by 42 is actually an element of the set A. And this tells us that the intersection B and C also belongs to the set A. So together we have that the set given by the intersection of B and C is actually equal to the set A. And then we can compute the size of B together with C since their intersection is the set A, that's all we need. This is equal to the size of B plus the size of C minus the size of their intersection. We have all these numbers, we have that B contains 30 elements, C contains 20 and their intersection being exactly A contains 10. So this comes out to 40. And we don't really need to minimize or maximize anything, the union of B and C is exactly 40. So the answer comes out to B.

marbles

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intersection

minimization