2018 AMC 12A Problems/Problem 6
Contents
Problem
For positive integers and
such that
, both the mean and the median of the set
are equal to
. What is
?
Solution 1
The mean and median are so
and
. Solving this gives
for
. (trumpeter)
Solution 2
You can immediately notice that the median is the average of
and
. There fore,
, so now we know we just are looking for
, which must be odd. This leaves our two remaining options, {(B)}21\qquad\textbf and {(D)}23\qquad\textbf. Note that if the answer is
, then
is odd, while the opposite is true for
if we get
. This fact will come in handy later on and prove itself useful when we solve for its parity. Since the average of the set of six numbers
is an integer, the sum of the terms must be even.
is odd by definition, so we know that
must also be odd, thus with some simple calculations
is odd. As with the previous few observations, we have eliminated all other answers, and
is the only remaining possibility left. Therefore
.
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 12 Problems and Solutions |
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