A woman named Alice, a distingushed astronomer, along with a group of her students, discovered a planet with a number of moons orbiting it. Alice decided to ignore the planet and concentrate on analyzing these moons.
She found that she could divide the moons equally among herself and her students, so that each person would have the same number of moons to study.
But her most brilliant student, Phil, suddenly suggested that the moons be broken up by families rather than individuals, so that people of the same family would work on a group of moons together. The only family relationships among these people were 3 distinct pairs of sisters. When the moons were divided along family lines, it was again possible to divide them so that each family group (each family group consisting of at least one person) got the same number of moons.
Before any study could commence, however, Pierre, another of Alice's students, decided not to study these moons, but to concentrate on the planet itself. With Pierre's decision, Alice at once dropped the division by family scheme, and divided the moons evenly among all the remaining people, including herself.
Two questions: What is the fewest number of students Alice could have had, and the fewest number of moons, to satisfy the requirements of this story ?
(In reply to
Solution by Penny)
I don't think Pierre and Phil are girls, Penny--unless they happen to have unusually morbid parents, a theory I think we can dismiss.
Anyway, on to my answer. And it's large, folks. Really large. Here it goes:
If we know that 6 students are girls (one of which is Alice) and at least two other ones must be Phil and Pierre, then the smallest amount of students the class can have is obviously 8. However, the smallest amount of moons is 72 (9 x 6 x 8)... Hmmm, the problem is kind of vague in what it's after.
If we know that by families, the number of moons must be divisible by 5, and if Pierre is absent it must be divisible by 7 the smallest number of moons there can be is 8 (total number of students) x 5 (families) x 7 (students minus Pierre) = 280, which seems pretty big to me.
Just read Charlie's answer. Glad to see I was on the right track.
Edited on March 30, 2004, 9:55 pm
re: Solution
