Difference between revisions of "2000 AMC 10 Problems/Problem 14"

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==Problem==
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#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==
 
 
 
71, 76, 80, 82, 91.
 
 
 
The sum of the first 2 must be even, so we must choose 2 evens or the 2 odds.
 
 
 
Let us look at the numbers (mod 3).
 
 
 
2,1,2,1,1.
 
 
 
If we choose the two odds, the next number must be a multiple of 3, of which there is none.
 
 
 
Similarly, if we choose 76,80 or 80,82, the next number must be a multiple of 3, of which there is none.
 
 
 
So we choose 76,82 first.
 
 
 
The next number must be 1 (mod 3), of which only 91 remains.
 
 
 
The sum of these is 249. This is equal to 1 (mod 4).
 
 
 
Thus, we need to choose one number that is 3 (mod 4). 71 is the only one that works.
 
 
 
Thus, 80 is the last score entered.
 
 
 
C.
 
 
 
==See Also==
 
 
 
{{AMC10 box|year=2000|num-b=13|num-a=15}}
 

Latest revision as of 23:50, 26 November 2011