Difference between revisions of "2010 AMC 12B Problems/Problem 14"
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− | == Problem | + | == Problem == |
Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> be positive integers with <math>a+b+c+d+e=2010</math> and let <math>M</math> be the largest of the sum <math>a+b</math>, <math>b+c</math>, <math>c+d</math> and <math>d+e</math>. What is the smallest possible value of <math>M</math>? | Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> be positive integers with <math>a+b+c+d+e=2010</math> and let <math>M</math> be the largest of the sum <math>a+b</math>, <math>b+c</math>, <math>c+d</math> and <math>d+e</math>. What is the smallest possible value of <math>M</math>? | ||
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− | However, all five numbers must be positive, | + | However, all five numbers must be positive, and the closest we can get to this is |
+ | |||
(<math>668, 3, 668, 3, 668</math>) | (<math>668, 3, 668, 3, 668</math>) | ||
+ | |||
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>. | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=13|num-a=15|ab=B}} | {{AMC12 box|year=2010|num-b=13|num-a=15|ab=B}} | ||
+ | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:38, 17 April 2024
Problem
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, and 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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