2014 AMC 10B Problems/Problem 18: Difference between revisions
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==Solution 2== | ==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> | 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>. Now, 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 + x2 = 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}} | ||
Revision as of 22:28, 24 October 2020
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 
. Now, 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
