AMC12 2023 A

AMC12 2023 A · Q13

AMC12 2023 A · Q13. It mainly tests Ratios & proportions, Basic counting (rules of product/sum).

In a table tennis tournament, every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?

在一场乒乓球锦标赛中，每位选手与其他每位选手恰好对战一次。虽然右手选手是左手选手的两倍，但左手选手赢得的比赛数比右手选手多 $40\%$。（没有平局，也没有双手选手。）总共进行了多少场比赛？

(A) 15 15

(B) 36 36

(D) 48 48

(E) 66 66

Answer

Correct choice: (B)

正确答案：（B）

Solution

We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$, and since $l = 1.4r$, $g = 2.4r$. Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is $n(n-1)/2$, the sum of the first $n-1$ numbers. Now, setting 36 and 48 equal to the equation will show that two consecutive numbers must have a product of 72 or 96. Clearly, $72=8*9$, so the answer is $\boxed{\textbf{(B) }36}$.

总比赛数是左右手选手赢得比赛数的总和。设 $g = l + r$，且 $l = 1.4r$，则 $g = 2.4r$。$r$ 和 $g$ 均为整数，故 $g/2.4$ 必须为整数，即 $g$ 必须被 $12$ 整除，故选项只剩 B 和 D。锦标赛总比赛数公式为 $\frac{n(n-1)}{2}$。设 $36$ 和 $48$ 等于该公式，则需找连续整数积为 $72$ 或 $96$。显然 $72=8\times9$，故总比赛数为 $\boxed{\textbf{(B) }36}$。

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