Difference between revisions of "2010 AMC 12B Problems/Problem 14"

Revision as of 16:23, 31 January 2021

Problem 14

Let $a$ , $b$ , $c$ , $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ , $b+c$ , $c+d$ and $d+e$ . What is the smallest possible value of $M$ ?

$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$

Solution 1

We want to try make $a+b$ , $b+c$ , $c+d$ , and $d+e$ as close as possible so that $M$ , the maximum of these, is smallest.

Notice that $2010=670+670+670$ . In order to express $2010$ as a sum of $5$ 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): $2010=670+1+670+1+668$ or $2010=670+1+669+1+669$ . We see that in both cases, the value of $M$ is $671$ , so the answer is $671 \Rightarrow \boxed{B}$ .

Solution 2

Since $a + b \le M$ , $d + e \le M$ , and $c < b + c \le M$ , we have that $2010 = a + b + c + d + e < 3M$ . Hence, $M > 670$ , or $M \ge 671$ .

For the values $(a,b,c,d,e) = (669,1,670,1,669)$ , $M = 671$ , so the smallest possible value of $M$ is $\boxed{671}$ . 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 $\boxed{B}$ .

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

Page

Toolbox

Search