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| − | ==Problem==
| + | #redirect [[2002 AMC 12A Problems/Problem 15]] |
| − | 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
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| − | <math>\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15</math>
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| − | ==Solution==
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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.
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| − | ==See Also==
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| − | {{AMC10 box|year=2002|ab=A|num-b=20|num-a=22}}
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| − | [[Category:Introductory Algebra Problems]] | |
