Problem 7 · Bounding an extreme by ordering · 5 points
Made by AI and not reviewed by a real person.
Five friends share 50 marbles, each getting a different positive number. The biggest share is twice the smallest; what is the largest possible middle share when the five shares are put in order?
- 12
- 10
- 9
- 14
- 11
Show a hint
- Put the shares in order. Making the middle share bigger can force changes at both ends.
- Try a middle share of 11. How small could all five shares be while following every rule?
Show the answer and solution
Answer: B — 10
If the middle share were 11 or more, the two larger shares would be at least 12 and 13. Since the largest is twice the smallest, the smallest must be at least 7, making the largest at least 14. The five shares would total at least 7 + 8 + 11 + 12 + 14 = 52—too many. But 7, 8, 10, 11, and 14 total 50 and follow every rule, so 10 is possible and is the greatest. The tempting 11 overlooks how the rule about the largest and smallest shares pushes the total above 50.