Difference between revisions of "2002 AMC 12P Problems/Problem 3"
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== Solution 1== | == Solution 1== | ||
− | Given an arbitrary product and an arbitrary amount of | + | 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.) 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}} |
Revision as of 18:07, 1 July 2024
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.) 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.