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 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?

答案

Solution:

Say the first monkey takes $8x$ bananas, keeping $6x$ and giving $x$ to each of the others; the second takes $8y,$ keeping $2y$ and giving $3y$ to each; the third takes $24z,$ keeping $2z$ and giving $11z$ to each. All divisions are whole numbers exactly when $x,$ $y,$ $z$ are positive integers. The final amounts are $6x + 3y + 11z,$ $x + 2y + 11z,$ and $x + 3y + 2z.$

The ratio $3 : 2 : 1$ says the first amount is triple the third and the second is double the third: $6x + 3y + 11z = 3(x + 3y + 2z) \quad\Longrightarrow\quad 3x + 5z = 6y,$ $x + 2y + 11z = 2(x + 3y + 2z) \quad\Longrightarrow\quad x + 4y = 7z.$ Substituting $x = 7z - 4y$ into the first equation gives $26z = 18y,$ so $9y = 13z.$ Thus $y = 13n$ and $z = 9n$ for a positive integer $n,$ and then $x = 63n - 52n = 11n.$

The total is $8x + 8y + 24z = (88 + 104 + 216)n = 408n,$ least when $n = 1:$ the answer is $408.$

2004 AIME II Problem 6

Solution:

Problem 6 in Other Years