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

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</math>
 
</math>
  
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== Solution ==
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== Solution 1 ==
  
 
As the unique mode is <math>8</math>, there are at least two <math>8</math>s.
 
As the unique mode is <math>8</math>, there are at least two <math>8</math>s.
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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.
 
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.
 
The solution that works is <math>(6,6,6,8,8,8,8,14)</math>
 
The solution that works is <math>(6,6,6,8,8,8,8,14)</math>
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==Solution 2 ==
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Let the 8 numbers be <math>a, b, c, d, e, f, g, h</math>, arranged in increasing order. Since the range of the eight numbers is 8, <math>h=a+8</math>.
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I claim that <math>d</math>,<math>e</math> must both be <math>8</math>. Since the median is 8, the mean of <math>d</math> and <math>e</math> must be 8. Let's assume that <math>d</math> and <math>e</math> aren't <math>8</math>. The mode of the collection is <math>8</math>, and if <math>d</math> and <math>e</math> aren't, then <math>8</math> must be between <math>d</math> and <math>e</math> (i.e. not in the collection). This is a contradiction, so <math>d</math> and <math>e</math> have to be 8.
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Now, we have the eight numbers are <math>a, b, c, 8, 8, f, g, a+8</math>.
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Since the mean is <math>8</math>, we have <math>a+b+c+8+8+f+g+a+8=8 \times 8 = 64</math>, giving us <math>2a+b+c+f+g=40</math>.
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Since <math>f, g \ge 8</math>, <math>f+g \ge 16</math>. Plugging that in, we have <math>2a+b+c \le 24</math>. Note that we can't do the same for <math>b</math> and <math>c</math>, but we can do <math>b+c \ge 2a</math>, giving us <math>4a \le 24</math>, which means <math>a\le 6</math>.
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We want to find <math>h=a+8\le 6+8=14</math>
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This is our answer, so <math>\boxed{(D)14}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 18:35, 10 August 2021

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 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. The solution that works is $(6,6,6,8,8,8,8,14)$

Solution 2

Let the 8 numbers be $a, b, c, d, e, f, g, h$, arranged in increasing order. Since the range of the eight numbers is 8, $h=a+8$.

I claim that $d$,$e$ must both be $8$. Since the median is 8, the mean of $d$ and $e$ must be 8. Let's assume that $d$ and $e$ aren't $8$. The mode of the collection is $8$, and if $d$ and $e$ aren't, then $8$ must be between $d$ and $e$ (i.e. not in the collection). This is a contradiction, so $d$ and $e$ have to be 8.

Now, we have the eight numbers are $a, b, c, 8, 8, f, g, a+8$.

Since the mean is $8$, we have $a+b+c+8+8+f+g+a+8=8 \times 8 = 64$, giving us $2a+b+c+f+g=40$.

Since $f, g \ge 8$, $f+g \ge 16$. Plugging that in, we have $2a+b+c \le 24$. Note that we can't do the same for $b$ and $c$, but we can do $b+c \ge 2a$, giving us $4a \le 24$, which means $a\le 6$. We want to find $h=a+8\le 6+8=14$ This is our answer, so $\boxed{(D)14}$

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

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