Question

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 the pile starts out with a positive number of bananas, 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 $3:2:1$, what is the least possible total for the number of bananas?

Answer 1

Answer:

408

Step-by-step explanation:

let the number of banas the first monkey takes be 8x. So, the first monkey will have 6x and x will go to each of the other monkeys. let the number of banaas the second monkey takes be 8y. It keeps 2y bananas and gives 3y to each other monkey. using the same way above, we get 6*68 = 408 bananas

Answer 2

Answer:

the answer is 408.

Step-by-step explanation: