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

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== Solution ==
 
== Solution ==
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If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
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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>
  
 
== 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}}

Revision as of 21:06, 30 December 2023

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

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 $1$ 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 $AM-GM$ 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 $2002= 2 \cdot 7 \cdot 11 \cdot 13$. 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 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

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