2018 AMC 8 第 13 题

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13.

Laila 参加了五次数学测试,每次满分 100100 分。每次成绩都是 00100100 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 8282。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 100100 points. Laila's score on each test was an integer between 00 and 100,100, 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 82.82. How many values are possible for Laila's score on the last test?

44

55

99

1010

1818

答案:A
知识点:平均数模运算
难度评级:1250
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文字解答:

因为五次测试平均分是 8282,所以五次总分为 582=4105\cdot82 = 410

设前四次测试的分数为 ff,最后一次测试的分数为 ll

我们知道 且 又因为 410=4f+l<5l410 = 4f + l < 5l ,所以 4105=82<l\frac{410}{5} = 82 < l f<l100f < l \leq 100 4f+l=410.4f + l = 410.

另外,由于 4f+l=4104f + l = 410 ,而 410410 除以 44 余下的部分必须来自最后一次分数。所以 ll 除以 44 也必须余 22,因为 4f4f 能被 44 整除。等价地, 因为 82<l10082 < l \leq 100l2mod4l \equiv 2 \mod 4,所以 ll 只能是 86,90,94,9886,90,94,98。这给出四组不同的解: 因此共有 44 种可能,正确答案是 Al2mod4.l \equiv 2 \mod 4 . (f,l)=(81,86);(80,90);(79,94);(78,98)\begin{align*} (f,l) =& (81,86);\\ &(80,90);\\ &(79,94);\\ &(78,98) \end{align*}

Since the average score on the five tests is 82,82, the total score of those five tests must be 582=410.5\cdot82 = 410 .

Now, let ff be the score on the first 4 tests and let ll be the score for the last test.

We know that f<l100f < l \leq 100 and 4f+l=410.4f + l = 410. And as 410=4f+l<5l,410 = 4f + l < 5l , we know 4105=82<l.\frac{410}{5} = 82 < l .

Also, since 4f+l=410,4f + l = 410 , and dividing 410410 by 44 gives us a remainder of 2, we know that dividing ll by 44 must leave a remainder of 22 as 4f4f will leave no remainder when divided by 4.4. Equivalently: l2mod4.l \equiv 2 \mod 4 . Since 82<l10082 < l \leq 100 and l2mod4,l \equiv 2 \mod 4, the only options for ll are 86,90,94,98.86,90,94,98. This yields four distinct solutions as follows: (f,l)=(81,86);(80,90);(79,94);(78,98)\begin{align*} (f,l) =& (81,86);\\ &(80,90);\\ &(79,94);\\ &(78,98) \end{align*} Therefore, there are 44 solutions, and A is the correct answer.

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