Difference between revisions of "2002 AMC 12P Problems/Problem 3"
(→Solution) |
(→Solution) |
||
| (2 intermediate revisions by the same user not shown) | |||
| Line 4: | Line 4: | ||
<math>\text{(A) }36 \qquad \text{(B) }38 \qquad \text{(C) }42 \qquad \text{(D) }44 \qquad \text{(E) }92</math> | <math>\text{(A) }36 \qquad \text{(B) }38 \qquad \text{(C) }42 \qquad \text{(D) }44 \qquad \text{(E) }92</math> | ||
| − | == Solution == | + | == Solution 1== |
| − | Given an arbitrary product and an arbitrary amount of terms to multiply to get that product, to maximize the sum, make all of the terms <math>1</math> with the last one being the number. To minimize the sum, make all of the terms equal to each other. (This is a corollary from the <math>AM-GM</math> proof.) Since 2002 is not a perfect cube, we have to make the terms as close to each other as possible. A good rule of thumb is the memorize the prime factorization of the AMC year that you are doing, which is <math>2002= 2 \cdot 7 \cdot 11 \cdot 13</math>. The three terms that are closest to each other that multiply to <math>2002</math> are <math>11, 13,</math> and <math>14</math>, so our answer is <math>11+13+14=\boxed{\textbf{B } 38}</math> | + | Given an arbitrary product and an arbitrary amount of terms to multiply to get that product, to maximize the sum, make all of the terms <math>1</math> with the last one being the number. To minimize the sum, make all of the terms equal to each other. (This is a corollary that follows from the <math>AM-GM</math> proof.) Since <math>2002</math> is not a perfect cube, we have to make the terms as close to each other as possible. A good rule of thumb is the memorize the prime factorization of the AMC year that you are doing, which is <math>2002= 2 \cdot 7 \cdot 11 \cdot 13</math>. The three terms that are closest to each other that multiply to <math>2002</math> are <math>11, 13,</math> and <math>14</math>, so our answer is <math>11+13+14=\boxed{\textbf{(B) } 38}</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=P|num-b=2|num-a=4}} | {{AMC12 box|year=2002|ab=P|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 22:07, 30 December 2023
Problem
The dimensions of a rectangular box in inches are all positive integers and the volume of the box is 
in
. Find the minimum possible sum of the three dimensions.
Solution 1
Given an arbitrary product and an arbitrary amount of terms to multiply to get that product, to maximize the sum, make all of the terms 
with the last one being the number. To minimize the sum, make all of the terms equal to each other. (This is a corollary that follows from the 
proof.) Since is not a perfect cube, we have to make the terms as close to each other as possible. A good rule of thumb is the memorize the prime factorization of the AMC year that you are doing, which is 
. The three terms that are closest to each other that multiply to are 
and 
, so our answer is 
See also
| 2002 AMC 12P (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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. 
