|
|
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #REDIRECT [[2000 AMC 12 Problems/Problem 9]] |
| − | | |
| − | 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 <math>71</math>, <math>76</math>, <math>80</math>, <math>82</math>, and <math>91</math>. What was the last score Mrs. Walter entered?
| |
| − | | |
| − | <math>\mathrm{(A)}\ 71 \qquad\mathrm{(B)}\ 76 \qquad\mathrm{(C)}\ 80 \qquad\mathrm{(D)}\ 82 \qquad\mathrm{(E)}\ 91</math>
| |
| − | | |
| − | ==Solution==
| |
| − | | |
| − | <math>71, 76, 80, 82, 91</math>
| |
| − | | |
| − | The sum of the first <math>2</math> scores must be even, so we must choose <math>2</math> evens or the <math>2</math> odds to be the first two scores.
| |
| − | | |
| − | Let us look at the numbers in mod <math>3</math>.
| |
| − | | |
| − | <math>2,1,2,1,1</math>
| |
| − | | |
| − | If we choose the two odds, the next number must be a multiple of <math>3</math>, of which there is none.
| |
| − | | |
| − | Similarly, if we choose <math>76,80</math> or <math>80,82</math>, the next number must be a multiple of <math>3</math>, of which there is none.
| |
| − | | |
| − | So we choose <math>76,82</math> first.
| |
| − | | |
| − | The next number must be 1 in mod 3, of which only <math>91</math> remains.
| |
| − | | |
| − | The sum of the first three scores is <math>249</math>. This is equivalent to <math>1</math> in mod <math>4</math>.
| |
| − | | |
| − | Thus, we need to choose one number that is <math>3</math> in mod <math>4</math>. <math>71</math> is the only one that works.
| |
| − | | |
| − | Thus, <math>80</math> is the last score entered.
| |
| − | | |
| − | <math>\boxed{\text{C}}</math>
| |
| − | | |
| − | ==See Also==
| |
| − | | |
| − | {{AMC10 box|year=2000|num-b=13|num-a=15}}
| |
