|
|
| Line 1: |
Line 1: |
| − | == Problem ==
| + | НИГЕР |
| − | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>A, M,</math> and <math>C</math> be [[nonnegative integer]]s such that <math>A + M + C=12</math>. What is the maximum value of <math>A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
| |
| − | | |
| − | <math> \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 } </math>
| |
| − | | |
| − | == Solution 1 ==
| |
| − | It is not hard to see that
| |
| − | <cmath>(A+1)(M+1)(C+1)=</cmath>
| |
| − | <cmath>AMC+AM+AC+MC+A+M+C+1</cmath>
| |
| − | Since <math>A+M+C=12</math>, we can rewrite this as
| |
| − | <cmath>(A+1)(M+1)(C+1)=</cmath>
| |
| − | <cmath>AMC+AM+AC+MC+13</cmath>
| |
| − | So we wish to maximize
| |
| − | <cmath>(A+1)(M+1)(C+1)-13</cmath>
| |
| − | Which is largest when all the factors are equal (consequence of AM-GM). Since <math>A+M+C=12</math>, we set <math>A=M=C=4</math>
| |
| − | Which gives us
| |
| − | <cmath>(4+1)(4+1)(4+1)-13=112</cmath>
| |
| − | so the answer is <math>\boxed{\text{E}}</math>.
| |
| − | I wish you understand this problem and can use it in other problems.
| |
| − | | |
| − | == Solution 2 (Nonrigorous) ==
| |
| − | | |
| − | If you know that to maximize your result you <math>\textit{usually}</math> have to make the numbers as close together as possible, (for example to maximize area for a polygon make it a square) then you can try to make <math>A,M</math> and <math>C</math> as close as possible. In this case, they would all be equal to <math>4</math>, so <math>AMC+AM+AC+MC=64+16+16+16=112</math>, giving you the answer of <math>\boxed{\text{E}}</math>.
| |
| − | | |
| − | == Solution 3 ==
| |
| − | | |
| − | Assume <math>A</math>, <math>M</math>, and <math>C</math> are equal to <math>4</math>. Since the resulting value of <math>AMC+AM+AC+MC</math> will be <math>112</math> and this is the largest answer choice, our answer is <math>\boxed{\textbf{(E) }112}</math>.
| |
| − | | |
| − | == Video Solution ==
| |
| − | https://youtu.be/lxqxQhGterg
| |
| − | | |
| − | == See also ==
| |
| − | {{AMC12 box|year=2000|num-b=11|num-a=13}}
| |
| − | | |
| − | [[Category:Introductory Algebra Problems]]
| |
| − | {{MAA Notice}}
| |
