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Difference between revisions of "2002 AMC 12A Problems/Problem 15"

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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.
 
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 we have 2 8's we can add find the numbers 1, 2, 3, 8, 8, 13, 14, 15.
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They add to 64 and 64/8 = 8.
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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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We can also see that they satisfy the need for the mode and median to be 8. This means that the answer will be 15
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<math>\boxed{\text{(D)}\ 15 }</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.
 
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The solution that works is <math>(6,6,6,8,8,8,8,14)</math>
 
  
 
== See Also ==
 
== See Also ==

Revision as of 00:20, 14 January 2022

The following problem is from both the 2002 AMC 12A #15 and 2002 AMC 10A #21, so both problems redirect to this page.

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 1

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 we have 2 8's we can add find the numbers 1, 2, 3, 8, 8, 13, 14, 15. They add to 64 and 64/8 = 8. We can also see that they satisfy the need for the mode and median to be 8. This means that the answer will be 15 $\boxed{\text{(D)}\ 15 }$.

See Also

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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