Difference between revisions of "2010 AMC 12B Problems/Problem 14"
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==Solution 3== | ==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.) | 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) | + | 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" = 0, 2010 will be divided into either 3 or 2 numbers for cases 1 and 2, respectively. Obviously dividing by 3 will yield a lower number, so we only have to consider case one. | ||
Dividing 2010 by 3 yields 670, or (670, 0, 670, 0, 670). However, all five numbers must be positive, therefore the closest we can get to this is (668, 3, 668, 3, 668). The lowest possible sum of two adjacent numbers then becomes 671, or <math>\boxed{B}</math>. | Dividing 2010 by 3 yields 670, or (670, 0, 670, 0, 670). However, all five numbers must be positive, therefore the closest we can get to this is (668, 3, 668, 3, 668). The lowest possible sum of two adjacent numbers then becomes 671, or <math>\boxed{B}</math>. | ||
Revision as of 16:22, 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" = 0, 2010 will be divided into either 3 or 2 numbers for cases 1 and 2, respectively. Obviously dividing by 3 will yield a lower number, so we only have to consider case one.
Dividing 2010 by 3 yields 670, or (670, 0, 670, 0, 670). However, all five numbers must be positive, therefore the closest we can get to this is (668, 3, 668, 3, 668). The lowest possible sum of two adjacent numbers then becomes 671, 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
