Difference between revisions of "2002 AMC 12P Problems/Problem 3"
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| + | {{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #3]] and [[2002 AMC 10P Problems|2002 AMC 10P #7]]}} | ||
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== 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. | ||
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<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 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.) | |
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| + | 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 
in
. Find the minimum possible sum of the three dimensions.
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 
proof.)
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 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. 
