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 $\text{(B)}$

See Also

1995 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AJHSME/AMC 8 Problems and Solutions
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