Difference between revisions of "1994 AHSME Problems/Problem 11"

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==Problem==
 
==Problem==
Three cubes of volume <math>1, 8</math> and <math>27</math> are glued together at their faces. The smallest possible surface area of the resulting configuration is
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Three cubes of volume <math>1</math>, <math>8</math> and <math>27</math> are glued together at their faces. The smallest possible surface area of the resulting configuration is
  
 
<math> \textbf{(A)}\ 36 \qquad\textbf{(B)}\ 56 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 72 \qquad\textbf{(E)}\ 74 </math>
 
<math> \textbf{(A)}\ 36 \qquad\textbf{(B)}\ 56 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 72 \qquad\textbf{(E)}\ 74 </math>
==Solution==
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==See Also==
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{{AHSME box|year=1994|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 20:50, 22 February 2024

Problem

Three cubes of volume $1$, $8$ and $27$ are glued together at their faces. The smallest possible surface area of the resulting configuration is

$\textbf{(A)}\ 36 \qquad\textbf{(B)}\ 56 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 72 \qquad\textbf{(E)}\ 74$

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 26 27 28 29 30
All AHSME Problems and Solutions

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