Difference between revisions of "2012 AMC 8 Problems/Problem 22"

(Solution 2)
(Solution 2)
Line 50: Line 50:
 
==Solution 2==
 
==Solution 2==
 
For those who don't like bashing problems out, use this method.
 
For those who don't like bashing problems out, use this method.
2,3,4,6,9,14,x,y,z
+
2, 3, 4, 6,9, 14, x, y, z
  
 
if we put x, y, z in front of 2, 3 can be the median
 
if we put x, y, z in front of 2, 3 can be the median

Revision as of 01:34, 3 November 2017

Problem

Let $R$ be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of $R$ ?

$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}8$




Solution 1

First, we find that the minimum value of the median of $R$ will be $3$.

We then experiment with sequences of numbers to determine other possible medians.

Median: $3$

Sequence: $-2, -1, 0, 2, 3, 4, 6, 9, 14$

Median: $4$

Sequence: $-1, 0, 2, 3, 4, 6, 9, 10, 14$

Median: $5$

Sequence: $0, 2, 3, 4, 5, 6, 9, 10, 14$

Median: $6$

Sequence: $0, 2, 3, 4, 6, 9, 10, 14, 15$

Median: $7$

Sequence: $2, 3, 4, 6, 7, 8, 9, 10, 14$

Median: $8$

Sequence: $2, 3, 4, 6, 8, 9, 10, 14, 15$

Median: $9$

Sequence: $2, 3, 4, 6, 9, 14, 15, 16, 17$

Any number greater than $9$ also cannot be a median of set $R$.

There are then $\boxed{\textbf{(D)}\ 7}$ possible medians of set $R$.


Solution 2

For those who don't like bashing problems out, use this method. 2, 3, 4, 6,9, 14, x, y, z

if we put x, y, z in front of 2, 3 can be the median

if we put x, y, z after 14, 9 is the median

therefore, 3,4,6,9,x,y,z can all be the medians

There are then $\boxed{\textbf{(D)}\ 7}$ possible medians of set $R$.

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png