Difference between revisions of "2014 AMC 10B Problems/Problem 18"
(→Solution 1) |
Mathlover66 (talk | contribs) (→Solution 2) |
||
Line 14: | Line 14: | ||
==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 | + | 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. |
==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}} |
Latest revision as of 23:47, 6 September 2020
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 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.
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.