1995 AJHSME Problems/Problem 21

Problem

A plastic snap-together cube has a protruding snap on one side and receptacle holes on the other five sides as shown. What is the smallest number of these cubes that can be snapped together so that only receptacle holes are showing?

$[asy] draw((0,0)--(4,0)--(4,4)--(0,4)--cycle); draw(circle((2,2),1)); draw((4,0)--(6,1)--(6,5)--(4,4)); draw((6,5)--(2,5)--(0,4)); draw(ellipse((5,2.5),0.5,1)); fill(ellipse((3,4.5),1,0.25),black); fill((2,4.5)--(2,5.25)--(4,5.25)--(4,4.5)--cycle,black); fill(ellipse((3,5.25),1,0.25),black); [/asy]$

$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution

It is rather easy to see that the only way that you can possibly have three cubes without protruding snaps is to have them in a triangle formation. However, in this case, the exterior angles are $120^\circ$ , whereas the outside angle of the cubes are $90^\circ$ each. A square like formation will work in the case of 4 cubes, so the answer is $\text{(B)}$ .

1995 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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
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1995 AJHSME Problems/Problem 21

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