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

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

Revision as of 22:04, 10 January 2009

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. Walter entered?

$\mathrm{(A)}\ 71 \qquad\mathrm{(B)}\ 76 \qquad\mathrm{(C)}\ 80 \qquad\mathrm{(D)}\ 82 \qquad\mathrm{(E)}\ 91$

Solution

$71, 76, 80, 82, 91$

The sum of the first $2$ scores must be even, so we must choose $2$ evens or the $2$ odds to be the first two scores.

Let us look at the numbers in 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 in mod 3, of which only $91$ remains.

The sum of the first three scores is $249$. This is equivalent to $1$ in mod $4$.

Thus, we need to choose one number that is $3$ in mod $4$. $71$ is the only one that works.

Thus, $80$ is the last score entered.

$\boxed{\text{C}}$

See Also

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
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