Difference between revisions of "2000 AMC 12 Problems/Problem 9"
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===Solution 3=== | ===Solution 3=== | ||
| − | If we add up all the numbers given: 71,76,80,82,91, we get 400. Before Mrs. Walter entered in the last number the sum had to be divisible by 4. So 400 minus the last number must be a multiple of 4. 80 is the only possible number, so the answer is <math>\mathrm{C}</math> | + | If we add up all the numbers given: <math>71,76,80,82,91</math>, we get <math>400</math>. Before Mrs. Walter entered in the last number the sum had to be divisible by <math>4</math>. So <math>400</math> minus the last number must be a multiple of <math>4</math>. <math>80</math> is the only possible number, so the answer is <math>\mathrm{C}</math> |
== See also == | == See also == | ||
Revision as of 17:30, 5 August 2014
- The following problem is from both the 2000 AMC 12 #9 and 2000 AMC 10 #14, so both problems redirect to this page.
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
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 
, 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 
. 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 
Solution 2
We know the first sum of the first three numbers must be divisible by 3, so we write out all 5 numbers 
, 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 
.
Solution 3
If we add up all the numbers given: 
, we get 
. Before Mrs. Walter entered in the last number the sum had to be divisible by 
. So minus the last number must be a multiple of . 
is the only possible number, so the answer is
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 | |
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
