Difference between revisions of "2014 AMC 10B Problems/Problem 18"
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<math> \textbf {(A) } 24 \qquad \textbf {(B) } 30 \qquad \textbf {(C) } 31\qquad \textbf {(D) } 33 \qquad \textbf {(E) } 35</math> | <math> \textbf {(A) } 24 \qquad \textbf {(B) } 30 \qquad \textbf {(C) } 31\qquad \textbf {(D) } 33 \qquad \textbf {(E) } 35</math> | ||
− | ==Solution== | + | ==Solution 1== |
We start off with the fact that the median is <math>9</math>, so we must have <math>a, b, c, d, e, 9, f, g, h, i, j</math>, listed in ascending order. Note that the integers do not have to be distinct. | We start off with the fact that the median is <math>9</math>, so we must have <math>a, b, c, d, e, 9, f, g, h, i, j</math>, listed in ascending order. Note that the integers do not have to be distinct. | ||
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Next, consider the case where there are <math>3</math> occurrences of <math>8</math> in the list. Now, we can have two occurrences of another integer in the list. We try <math>1,1,8,8,8,9,9,10,10,11,j</math>. Following the same process as above, we get <math>j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35</math>. As this is the highest choice in the list, we know this is our answer. Therefore, the answer is <math>\boxed{\textbf{(E) }35}</math> | Next, consider the case where there are <math>3</math> occurrences of <math>8</math> in the list. Now, we can have two occurrences of another integer in the list. We try <math>1,1,8,8,8,9,9,10,10,11,j</math>. Following the same process as above, we get <math>j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35</math>. As this is the highest choice in the list, we know this is our answer. Therefore, the answer is <math>\boxed{\textbf{(E) }35}</math> | ||
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+ | ==Solution 2== | ||
+ | Note that <math>x_1 + \ldots + x_{11} = 110</math> let <math>x_6 = 9</math> so <math>x_1 + \ldots + x_5 + x_7 + \ldots + x_{11} = 101</math>. To maximize the value of <math>x_i</math> where <math>i</math> ranges from <math>1</math> to <math>11</math>, we let any <math>7</math> elements be <math>1,2,\ldots,7</math> so <math>x_1 + x_2 + x_3 = 57</math>. Now we have to let one of above <math>3</math> values = <math>8</math> hence <math>x_1 + x_2 = 49</math> now let <math>x_1 = 35</math>, <math>x_2 = 14</math> hence <math>\boxed{\textbf{(E) }35}</math> is the answer. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=17|num-a=19}} | {{AMC10 box|year=2014|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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Latest revision as of 16:44, 30 January 2021
Contents
Problem
A list of positive integers has a mean of , a median of , and a unique mode of . What is the largest possible value of an integer in the list?
Solution 1
We start off with the fact that the median is , so we must have , listed in ascending order. Note that the integers do not have to be distinct.
Since the mode is , we have to have at least occurrences of in the list. If there are occurrences of in the list, we will have . In this case, since is the unique mode, the rest of the integers have to be distinct. So we minimize in order to maximize . If we let the list be , then .
Next, consider the case where there are occurrences of in the list. Now, we can have two occurrences of another integer in the list. We try . Following the same process as above, we get . As this is the highest choice in the list, we know this is our answer. Therefore, the answer is
Solution 2
Note that let so . To maximize the value of where ranges from to , we let any elements be so . Now we have to let one of above values = hence now let , hence is the answer.
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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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