Difference between revisions of "2000 AMC 12 Problems/Problem 9"
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<math>\text{(A)} \ 71 \qquad \text{(B)} \ 76 \qquad \text{(C)} \ 80 \qquad \text{(D)} \ 82 \qquad \text{(E)} \ 91</math> | <math>\text{(A)} \ 71 \qquad \text{(B)} \ 76 \qquad \text{(C)} \ 80 \qquad \text{(D)} \ 82 \qquad \text{(E)} \ 91</math> | ||
− | + | == Solutions == | |
− | == Solution 1 == | + | === Solution 1 === |
− | |||
The first number is divisible by <math>1</math>. | The first number is divisible by <math>1</math>. | ||
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Since <math>80\equiv 2\pmod{3}</math>, the fourth number must be <math>2\pmod{3}</math>. That number is <math>71</math> and <math>71</math> only. The next number must be <math>91</math> since the sum of the first two numbers is even. So the only arrangement of the scores <math>76, 82, 91, 71, 80</math> <math>\Rightarrow \text{(C)}</math> | Since <math>80\equiv 2\pmod{3}</math>, the fourth number must be <math>2\pmod{3}</math>. That number is <math>71</math> and <math>71</math> only. The next number must be <math>91</math> since the sum of the first two numbers is even. So the only arrangement of the scores <math>76, 82, 91, 71, 80</math> <math>\Rightarrow \text{(C)}</math> | ||
− | == Solution 2 == | + | === Solution 2 === |
We know the first sum of the first three numbers must be divisible by <math>3,</math> so we write out all <math>5</math> numbers <math>\pmod{3}</math>, which gives <math>2,1,2,1,1,</math> respectively. Clearly, the only way to get a number divisible by <math>3</math> 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 <math>4, 71</math> must be next. That leaves <math>80</math> for last, so the answer is <math>\mathrm{C}</math>. | We know the first sum of the first three numbers must be divisible by <math>3,</math> so we write out all <math>5</math> numbers <math>\pmod{3}</math>, which gives <math>2,1,2,1,1,</math> respectively. Clearly, the only way to get a number divisible by <math>3</math> 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 <math>4, 71</math> must be next. That leaves <math>80</math> for last, so the answer is <math>\mathrm{C}</math>. | ||
− | == Video == | + | == Video Solution == |
− | https://www.youtube.com/watch?v=IJ4xXPEfrzc | + | https://www.youtube.com/watch?v=IJ4xXPEfrzc |
== See also == | == See also == |
Latest revision as of 13:34, 19 January 2021
- 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 , , , , and . What was the last score Mrs. Walters entered?
Solutions
Solution 1
The first number is divisible by .
The sum of the first two numbers is even.
The sum of the first three numbers is divisible by
The sum of the first four numbers is divisible by
The sum of the first five numbers is
Since is divisible by the last score must also be divisible by Therefore, the last score is either or
Case 1: is the last number entered.
Since , the fourth number must be divisible by but none of the scores are divisible by
Case 2: is the last number entered.
Since , the fourth number must be . That number is and only. The next number must be 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 so we write out all numbers , which gives respectively. Clearly, the only way to get a number divisible by 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 must be next. That leaves for last, so the answer is .
Video Solution
https://www.youtube.com/watch?v=IJ4xXPEfrzc
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.