AIME 1993 · 第 3 题

Solution 1

Suppose that the number of fish is $x$ and the number of contestants is $y$. The $y-(9+5+7)=y-21$ fishers that caught $3$ or more fish caught a total of $x - \left(0\cdot(9) + 1\cdot(5) + 2\cdot(7)\right) = x - 19$ fish. Since they averaged $6$ fish,

$6 = \frac{x - 19}{y - 21} \Longrightarrow x - 19 = 6y - 126.$

Similarily, those who caught $12$ or fewer fish averaged $5$ fish per person, so

$5 = \frac{x - (13(5) + 14(2) + 15(1))}{y - 8} = \frac{x - 108}{y - 8} \Longrightarrow x - 108 = 5y - 40.$

Solving the two equation system, we find that $y = 175$ and $x = \boxed{943}$, the answer.

Solution 2

Let $f$ be the total number of fish caught by the contestants who didn't catch $0, 1, 2, 3, 13, 14$, or $15$ fish and let $a$ be the number of contestants who didn't catch $0, 1, 2, 3, 13, 14$, or $15$ fish. From $\text{(b)}$, we know that $\frac{69+f+65+28+15}{a+31}=6\implies f=6a+9$. From $\text{(c)}$ we have $\frac{f+69+14+5}{a+44}=5\implies f=5a+132$. Using these two equations gets us $a=123$. Plug this back into the equation to get $f=747$. Thus, the total number of fish caught is $5+14+69+f+65+28+15=\boxed{943}$ - Heavytoothpaste

Solution 3

Let $x$ be the average number of fish caught among $n$ of those who caught $4$ to $12$ fishes. According to statement B:

$$\frac{xn + 3 \cdot 23 + 13 \cdot 5 + 14 \cdot 2 + 15}{n + 31} = 6$$ $$\implies xn - 6n = 9$$ According to statement C:

$$\frac{xn + 0 \cdot 9 + 1 \cdot 5 + 2 \cdot 7 + 3 \cdot 23}{n + 44} = 5$$ $$\implies xn - 5n = 132$$ We then solve for $n$ and then $xn$. Note $xn$ is the number of fish caught by the $n$ people we defined above.

$$n = 123$$ $$xn = 132 + 5(123) = 747$$ Thus, the total number of fish caught is

$$747 + 1 \cdot 5 + 2 \cdot 7 + 3 \cdot 23 + 13 \cdot 5 + 14 \cdot 2 + 15 \cdot 1 = \boxed{943}$$ ~Totient4Breakfast