Difference between revisions of "1985 AHSME Problems/Problem 20"
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==Problem== | ==Problem== | ||
− | A wooden | + | A wooden cube with edge length <math>n</math> units (where <math>n</math> is an integer <math>>2</math>) is painted black all over. By slices parallel to its faces, the cube is cut into <math>n^3</math> smaller cubes each of unit edge length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is <math>n</math>? |
<math> \mathrm{(A)\ } 5 \qquad \mathrm{(B) \ }6 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 8 \qquad \mathrm{(E) \ }\text{none of these} </math> | <math> \mathrm{(A)\ } 5 \qquad \mathrm{(B) \ }6 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 8 \qquad \mathrm{(E) \ }\text{none of these} </math> | ||
==Solution== | ==Solution== | ||
− | + | Observe that if we remove the outer layer of unit cubes from the entire cube, what remains is a smaller cube of side length <math>(n-2)</math>, which contains all of the unpainted cubes and no others. This shows that there are exactly <math>(n-2)^3</math> unpainted cubes. Similarly, taking one face of the cube and removing the outer edge leaves a square of side length <math>(n-2)</math> containing all of the cubes on that face with exactly one face painted. Making the same argument for the other <math>5</math> faces as well, we deduce that there are a total of <math>6(n-2)^2</math> cubes with only one face painted. | |
− | < | + | Accordingly, we require <cmath>\begin{align*}(n-2)^3 = 6(n-2)^2 &\iff n-2 = 6 \qquad \text{(as } n > 2\text{, so } n-2 \neq 0\text{)} \ &\iff n = \boxed{\text{(D)} \ 8}.\end{align*}</cmath> |
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==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=19|num-a=21}} | {{AHSME box|year=1985|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:17, 19 March 2024
Problem
A wooden cube with edge length units (where is an integer ) is painted black all over. By slices parallel to its faces, the cube is cut into smaller cubes each of unit edge length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is ?
Solution
Observe that if we remove the outer layer of unit cubes from the entire cube, what remains is a smaller cube of side length , which contains all of the unpainted cubes and no others. This shows that there are exactly unpainted cubes. Similarly, taking one face of the cube and removing the outer edge leaves a square of side length containing all of the cubes on that face with exactly one face painted. Making the same argument for the other faces as well, we deduce that there are a total of cubes with only one face painted.
Accordingly, we require
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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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