AIME 2004 II · 第 6 题

Solution 1

In this solution, you start out the same as solution one. Convert everything into the fractions of largest denominator terms (this is necessary) until you get

$27x=11z$ $27y=13z$

While solving, and make sure to leave a list of numbers that must divide $x$, $y$, and $z$. For example, just by looking at the basic fractions you receive from writing the starting equations, 24 divides $z$, 8 divides $y$ and $x$. In the expression above, it's also clear that 27 divides $z$, 13 divides $y$, and 11 divides $z$. You might be wondering why I wrote the expression in terms of z. That's because $z$ has the largest divisor. The LCM of all it's divisors shows that $z$ must be divisible by 216. The total amount of bananas can be found to equal to $17z/9$. This means there are two possible solutions under 1000: 408 and 816. Trial and error can be done quickly to find the smallest possible solution to be $\boxed{408}$.

-jackshi2006

Solution 2 (A lot of Thinking Required dEfInItELy )

Let $A,B,C$ be the fraction of bananas taken by the first, second, and third monkeys respectively. Then we have the system of equations

$$\frac{3}{4}A+\frac{3}{8}B+\frac{11}{24}C=\frac{1}{2}$$ $$\frac{1}{8}A+\frac{1}{4}B+\frac{11}{24}C=\frac{1}{3}$$ $$\frac{1}{8}A+\frac{3}{8}B+\frac{2}{24}C=\frac{1}{6}.$$ Solve this your favorite way to get that

$$(A,B,C)=\left( \frac{11}{51}, \frac{13}{51}, \frac{9}{17} \right).$$ We need the amount taken by the first and second monkeys to be divisible by 8 and the third by 24 (but for the third, we already have divisibility by 9). Thus our minimum is $8 \cdot 51 = \boxed{408}.$

~Dhillonr25