2010 AMC 12B Problems/Problem 14
Contents
[hide]Problem 14
Let ,
,
,
, and
be positive integers with
and let
be the largest of the sum
,
,
and
. What is the smallest possible value of
?
Solution 1
We want to try make ,
,
, and
as close as possible so that
, the maximum of these, is smallest.
Notice that . In order to express
as a sum of
numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible):
or
. We see that in both cases, the value of
is
, so the answer is
.
Solution 2
First, note that, simply by pigeonhole, at least one of a, b, c, d, e is greater than or equal to so none of C, D, or E can be the answer. Thus, the answer is A or B. We will show that A is unattainable, leaving us with B as the only possible answer.
Assume WLOG that is the largest sum. So
meaning
Because we let
we must have
and
Adding these inequalities gives
But we just showed that
which means that
a contradiction because we are told that all the variables are positive.
Therefore, the answer is
Solution 3
Since ,
, and
, we have that
. Hence,
, or
.
For the values ,
, so the smallest possible value of
is
. The answer is (B).
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.