Grades 11-12 Video Solutions 2011-11&12 Video Solutions 2011 problem20

11&12 Video Solutions 2011 problem20

Video Transcription

Video Summary

The problem involves minimizing the number of marbles in a bowl, labeled by numbers divisible by 6, 7, and 42. Using a Venn diagram, we see that: marbles divisible by 6 (30), divisible by 7 (20), and divisible by 42 (10). Since marbles divisible by 42 are divisible by both 6 and 7, they comprise the intersection of those sets. Thus, the number of marbles is calculated as: 30 (divisible by 6) + 20 (divisible by 7) - 10 (intersection/divisible by 42) = 40. Hence, the smallest possible number of marbles is 40.

Keywords

marbles

Venn diagram

divisible

intersection

minimization