Difference between revisions of "2000 AMC 12 Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | {{ | + | 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 <math>400\equiv 76\equiv 1\pmod{3}</math>, 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 <math>80\equiv 2\pmod{3}</math>, the fourth number must be 2mod3. 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 possible arrangements of the scores are 82, 76, 91, 71, 80 and 76, 82, 91, 71, 80 <math>\Rightarrow \text{(C)}</math> | ||
== See also == | == See also == | ||
Revision as of 15:11, 3 January 2008
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?
Solution
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 
, 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 
, the fourth number must be 2mod3. 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 possible arrangements of the scores are 82, 76, 91, 71, 80 and 76, 82, 91, 71, 80 
See also
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
