Difference between revisions of "2010 AMC 12B Problems/Problem 14"
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However, all five numbers must be positive, therefore the closest we can get to this is | However, all five numbers must be positive, therefore the closest we can get to this is | ||
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(<math>668, 3, 668, 3, 668</math>) | (<math>668, 3, 668, 3, 668</math>) | ||
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The lowest possible sum of two adjacent numbers then becomes <math>671</math>, or <math>\boxed{B}</math>. | The lowest possible sum of two adjacent numbers then becomes <math>671</math>, or <math>\boxed{B}</math>. |
Revision as of 17:25, 31 January 2021
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
Since ,
, and
, we have that
. Hence,
, or
.
For the values ,
, so the smallest possible value of
is
. The answer is (B).
~ math31415926535
Solution 3
Notice that only the sums of adjacent numbers matter. (For example, a & c could be extremely high, as long as b is relatively low.) Therefore creating "mountains" and "valleys" is the best way to lower the sum of adjacent numbers. We can do
1. (high, low, high, low, high)
or
2. (low, high, low, high, low)
In the extreme case that each "low" = ,
will be divided into either
or
numbers for cases 1 and 2, respectively. Obviously dividing by
will yield a lower number, so we consider case 1.
Dividing by
yields
, or
()
However, all five numbers must be positive, therefore the closest we can get to this is
()
The lowest possible sum of two adjacent numbers then becomes , or
.
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 |
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