Problem:
Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio , what is the least possible total for the number of bananas?
Solution:
Let the first monkey take bananas from the pile, keeping and giving to each of the others. Let the second monkey take bananas from the pile, keeping and giving to each of the others. Let the third monkey take bananas from the pile, keeping and giving to each of the others. The total number of bananas is . The given ratios imply that and . Simplify these equations to obtain and . Eliminate to obtain . Then and , where is a positive integer. Substitute to find that . Thus, the least possible values for and are and , respectively, and the least possible total is .
The problems on this page are the property of the MAA's American Mathematics Competitions