During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

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

m (Solution 2)
(Solution 1)
 
(3 intermediate revisions by 2 users not shown)
Line 25: Line 25:
 
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.
  
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.
+
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.
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>.
+
We can also see that they satisfy the need for the mode and median to be 8. This means that the answer will be
 
+
<math>\boxed{\text{(D)}\ 15 }</math>.
===Note===
 
 
 
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>
 
 
 
==Solution 2 ==
 
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>.
 
 
 
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.
 
 
 
Now, we have the eight numbers are <math>a, b, c, 8, 8, f, g, a+8</math>.
 
 
 
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>.
 
 
 
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>.
 
We want to find <math>h=a+8\le 6+8=14</math>
 
This is our answer, so <math>\boxed{\text{(D)}14}</math>
 
  
 
== See Also ==
 
== See Also ==

Latest revision as of 23:21, 13 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 $\boxed{\text{(D)}\ 15 }$.

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS