Difference between revisions of "2014 AMC 10B Problems/Problem 18"
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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 x1 + ... + x11 = 110 let x6 = 9 so x1 + ... + x5 + x7+ ... + x11 = 101 now to maximise the value of xi where i ranges from 1 to 11, we let any 7 elements be 1,2,...,7 so x1 + x2 + x3 = 57 now we have to let one of above 3 values = 8 hence x1 + x2 = 49 now let x1 = 35 , x2 =14 hence 35 is the answe | ||
==See Also== | ==See Also== | ||
Revision as of 02:02, 16 August 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
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:
Solution 2: Note that x1 + ... + x11 = 110 let x6 = 9 so x1 + ... + x5 + x7+ ... + x11 = 101 now to maximise the value of xi where i ranges from 1 to 11, we let any 7 elements be 1,2,...,7 so x1 + x2 + x3 = 57 now we have to let one of above 3 values = 8 hence x1 + x2 = 49 now let x1 = 35 , x2 =14 hence 35 is the answe
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
Solution 2: Note that x1 + ... + x11 = 110 let x6 = 9 so x1 + ... + x5 + x7+ ... + x11 = 101 now to maximise the value of xi where i ranges from 1 to 11, we let any 7 elements be 1,2,...,7 so x1 + x2 + x3 = 57 now we have to let one of above 3 values = 8 hence x1 + x2 = 49 now let x1 = 35 , x2 =14 hence 35 is the answer
