Question

Laila 参加了五次数学测试，每次满分 $100$ 分。每次成绩都是 $0$ 到 $100$ 之间的整数。她前四次测试得了相同分数，最后一次分数更高。五次测试的平均分是 $82$ 。Laila 最后一次测试的分数可能有多少个不同的值？

Laila took five math tests, each worth a maximum of $100$ points. Laila’s score on each test was an integer between $0$ and $100,$ inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was $82.$ How many values are possible for Laila’s score on the last test?

$4$

$5$

$9$

$10$

$18$

答案：A

知识点：平均数模运算

难度评级：1250

小提示：

设前四次相同分数为 $f$ ，最后一次分数为 $l$

Let the repeated score be $f$ and the last test score be $l$ .

大提示：

使用 $4f+l=410$ ，并注意 $l$ 大于平均分

Use $4f+l=410$ and the fact that $l$ is larger than the average.

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文字解答：

因为五次测试平均分是 $82$ ，所以五次总分为 $5\cdot82 = 410$ 。

设 $f$ 为前 $4$ 次测试的分数， $l$ 为最后一次测试的分数。

我们知道 $f < l \leq 100$ 且 $4f + l = 410\text{。}$ 又因为 $410 = 4f + l < 5l$ ，所以 $\frac{410}{5} = 82 < l$ 。

另外，由于 $4f + l = 410$ ，而 $410$ 除以 $4$ 的余数是 $2$ ，所以 $l$ 除以 $4$ 也必须余 $2$ ，因为 $4f$ 除以 $4$ 没有余数。等价地， $l \equiv 2 \mod 4\text{。}$ 因为 $82 < l \leq 100$ 且 $l \equiv 2 \mod 4$ ，所以 $l$ 只能是 $86,90,94,98$ 。这给出四组不同的解： $\begin{align*} (f,l) =& (81,86);\\ &(80,90);\\ &(79,94);\\ &(78,98) \end{align*}$ 因此共有 $4$ 种可能，正确答案是 A。

Since the average score on the five tests is $82,$ the total score of those five tests must be $5\cdot82 = 410 .$

Now, let $f$ be the score on the first $4$ tests and let $l$ be the score for the last test.

We know that $f < l \leq 100$ and $4f + l = 410.$ And as $410 = 4f + l < 5l ,$ we know $\frac{410}{5} = 82 < l .$

Also, since $4f + l = 410 ,$ and dividing $410$ by $4$ gives us a remainder of $2,$ we know that dividing $l$ by $4$ must leave a remainder of $2$ as $4f$ will leave no remainder when divided by $4.$ Equivalently: $l \equiv 2 \mod 4 .$ Since $82 < l \leq 100$ and $l \equiv 2 \mod 4,$ the only options for $l$ are $86,90,94,98.$ This yields four distinct solutions as follows: $\begin{align*} (f,l) =& (81,86);\\ &(80,90);\\ &(79,94);\\ &(78,98) \end{align*}$ Therefore, there are $4$ solutions, and A is the correct answer.

2018 AMC 8 第 13 题