专家免费评估留学案例雅思保分免费备考资料留学热线 400-880-6200

AMC10每日一题（2000年真题#14）

2018-04-20
727 人浏览
分享
收藏

2000 AMC 10 竞赛试题/第14题

Problem

Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71,76,80,82, and 91. What was the last score Mrs. Walters entered?

$\text{(A)} \ 71 \qquad \text{(B)} \ 76 \qquad \text{(C)} \ 80 \qquad \text{(D)} \ 82 \qquad \text{(E)} \ 91$

Solution

Solution 1

The first number is divisible by 1.

The sum of the first two numbers is even.

The sum of the first three numbers is divisible by 3.

The sum of the first four numbers is divisible by 4.

The sum of the first five numbers is 400.

Since 400 is divisible by 4, the last score must also be divisible by 4. Therefore, the last score is either 76 or 80.

Case 1: 76 is the last number entered.

Since $400\equiv 76\equiv 1\pmod{3}$ , the fourth number must be divisible by 3, but none of the scores are divisible by 3.

Case 2: 80 is the last number entered.

Since $80\equiv 2\pmod{3}$ , the fourth number must be $2\pmod{3}$ . That number is 71 and only 71. The next number must be 91, since the sum of the first two numbers is even. So the only arrangement of the scores $76, 82, 91, 71, 80$ $\Rightarrow \text{(C)}$

Solution 2

We know the first sum of the first three numbers must be divisible by 3, so we write out all 5 numbers $\pmod{3}$ , which gives 2,1,2,1,1, respectively. Clearly the only way to get a number divisible by 3 by adding three of these is by adding the three ones. So those must go first. Now we have an odd sum, and since the next average must be divisible by 4, 71 must be next. That leaves 80 for last, so the answer is $\mathrm{C}$ .