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Difference between revisions of "2002 AMC 10A Problems/Problem 21"

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==Solution==
 
==Solution==
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Since the mean is 8, the sum of the eight numbers must be 64. Now we use the fact that the range is 8 to experiment. Letting the smallest number be 7, we find the largest is 15. Since the unique mode is 8, there must be atleast two eights. The sum of those four numbers is 38 so the other four numbers sum to 26. That means the average of the other four numbers is 6.5, which is below 7. Hence, atleast one of the numbers is below 7, which contradicts our assumption that the smallest is 7.
 
  
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Now let's try with the smallest number equal to 6. The largest number is then 14. If we take two sixes and four eights, we get a sequence that fits all the requirements. Hence, our answer is <math>\boxed{\text{(D)}\ 14 }</math>.
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As the unique mode is <math>8</math>, there are at least two <math>8</math>s.
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As the range is <math>8</math> and one of the numbers is <math>8</math>, the largest one can be at most <math>16</math>.
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If the largest one is <math>16</math>, then the smallest one is <math>8</math>, and thus the mean is strictly larger than <math>8</math>, which is a contradiction.
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If the largest one is <math>15</math>, then the smallest one is <math>7</math>. This means that we already know four of the values: <math>8</math>, <math>8</math>, <math>7</math>, <math>15</math>. Since the mean of all the numbers is <math>8</math>, their sum must be <math>64</math>. Thus the sum of the missing four numbers is <math>64-8-8-7-15=26</math>. But if <math>7</math> is the smallest number, then the sum of the missing numbers must be at least <math>4\cdot 7=28</math>, which is again a contradiction.
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If the largest number is <math>14</math>, we can easily find the solution <math>(6,6,6,8,8,8,8,14)</math>. Hence, our answer is <math>\boxed{\text{(D)}\ 14 }</math>.
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===Note===
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The solution for <math>14</math> is, in fact, unique. As the median must be <math>8</math>, this means that both the <math>4^\text{th}</math> and the <math>5^\text{th}</math> number, when ordered by size, must be <math>8</math>s. This gives the partial solution <math>(6,a,b,8,8,c,d,14)</math>. For the mean to be <math>8</math> each missing variable must be replaced by the smallest allowed value.
  
 
==See Also==
 
==See Also==

Revision as of 13:38, 24 January 2009

Problem

The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is

$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solution

As the unique mode is $8$, there are at least two $8$s.

As the range is $8$ and one of the numbers is $8$, the largest one can be at most $16$.

If the largest one is $16$, then the smallest one is $8$, and thus the mean is strictly larger than $8$, which is a contradiction.

If the largest one is $15$, then the smallest one is $7$. This means that we already know four of the values: $8$, $8$, $7$, $15$. Since the mean of all the numbers is $8$, their sum must be $64$. Thus the sum of the missing four numbers is $64-8-8-7-15=26$. But if $7$ is the smallest number, then the sum of the missing numbers must be at least $4\cdot 7=28$, which is again a contradiction.

If the largest number is $14$, we can easily find the solution $(6,6,6,8,8,8,8,14)$. Hence, our answer is $\boxed{\text{(D)}\ 14 }$.

Note

The solution for $14$ is, in fact, unique. As the median must be $8$, this means that both the $4^\text{th}$ and the $5^\text{th}$ number, when ordered by size, must be $8$s. This gives the partial solution $(6,a,b,8,8,c,d,14)$. For the mean to be $8$ each missing variable must be replaced by the smallest allowed value.

See Also

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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