Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AMC 12P Problems/Problem 3"

(→‎Solution)
(→‎See also)
 
(5 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #3]] and [[2002 AMC 10P Problems|2002 AMC 10P #7]]}}
 +
 
== Problem ==
 
== Problem ==
 
The dimensions of a rectangular box in inches are all positive integers and the volume of the box is <math>2002</math> in<math>^3</math>. Find the minimum possible sum of the three dimensions.
 
The dimensions of a rectangular box in inches are all positive integers and the volume of the box is <math>2002</math> in<math>^3</math>. Find the minimum possible sum of the three dimensions.
Line 4: Line 6:
 
<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 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>
+
Given an arbitrary product and an arbitrary amount of positive integers to multiply to get that product, make the terms as close to each other as possible to minimize the sum. (This is a corollary that follows from the <math>AM-GM</math> proof.)  
 +
 
 +
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 ==
 +
{{AMC10 box|year=2002|ab=P|num-b=6|num-a=8}}
 
{{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 17:13, 14 July 2024

The following problem is from both the 2002 AMC 12P #3 and 2002 AMC 10P #7, so both problems redirect to this page.

Problem

The dimensions of a rectangular box in inches are all positive integers and the volume of the box is $2002$ in$^3$. Find the minimum possible sum of the three dimensions.

$\text{(A) }36 \qquad \text{(B) }38 \qquad \text{(C) }42 \qquad \text{(D) }44 \qquad \text{(E) }92$

Solution 1

Given an arbitrary product and an arbitrary amount of positive integers to multiply to get that product, make the terms as close to each other as possible to minimize the sum. (This is a corollary that follows from the $AM-GM$ proof.)

A good rule of thumb is the memorize the prime factorization of the AMC year that you are doing, which is $2002= 2 \cdot 7 \cdot 11 \cdot 13$. The three terms that are closest to each other that multiply to $2002$ are $11, 13,$ and $14$, so our answer is $11+13+14=\boxed{\textbf{(B) } 38}$.

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 10 Problems and Solutions
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12P_Problems/Problem_3&oldid=222369"