AMC10 2023 A

AMC10 2023 A · Q16

AMC10 2023 A · Q16. It mainly tests Percent, 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?

在一次乒乓球锦标赛中，每位参赛者与其他每位参赛者恰好对战一次。虽然右手球员的数量是左手球员的两倍，但左手球员赢得的比赛数量比右手球员多40%。（没有平局，也没有双手球员。）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 $\boxed{\textbf{(B) }36}$.

我们知道总比赛数是左手和右手球员赢得比赛数的总和。于是有 $g = l + r$，且 $l = 1.4r$，所以 $g = 2.4r$。由于 $r$ 和 $g$ 都是整数，$g/2.4$ 也必须是整数。从这里可以看出 $g$ 必须能被12整除，因此只有选项B和D。然后，我们知道锦标赛的总比赛数公式是 $n(n-1)/2$，即前 $n-1$ 个数的和。将36和48代入方程，会发现两个连续数的乘积必须是72或96。显然，$72=8\times9$，所以答案是 $\boxed{\textbf{(B)}36}$。

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