Difference between revisions of "2014 AMC 10B Problems/Problem 18"
(→Solution) |
(→Solution) |
||
Line 11: | Line 11: | ||
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> | ||
+ | |||
+ | solution by crastybow | ||
==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 19:39, 20 February 2014
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 by crastybow
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.