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High School POW: the cube
Posted:
Jul 17, 1997 11:29 AM


Given a painted cube n units on a side that is cut into smaller cubes 1 unit on a side, how many smaller cubes will have 3 painted surfaces, 2, 1, 0 painted surfaces?
answer: 8 cubes 3 painted surfaces (corners) 12(n2) cubes 2 pntd srfc (edges) 6(n2)^2 cbs 1 ps (interior exposed surfaces) (n2)^3 cbs 0 ps (interior
We approached the problem individually at first, then we began to compare notes. We first agreed that N could not be 1 because that cube would have 6 painted sides
Some started making a table for cubes of increasing dimension: 2*2*2, 3*3*3, 4*4*4, etc.They counted cubes in each of the four categories for each size cube, filled in the table and looked for a pattern.
Others took an algebraic approach:
if n was some number not equal to 1, the cube would have 8 corners with 3 ptd fcs;
each of the 12 edges would minus the corners would have (n2) cubes with ptd fcs;
each of the 6 exposed surfaces (minus corners and edges) would have (n2)^2 cubes with ptd fcs;
all the inner unpainted cubes would for a cube with side (n2) for a total of (n2)^3 small cubes.
The general form for the total num,ber of cubes works out to be: n^3  4n^2 + 18n  28
We plugged values from the table into the algebraic formula and found that both methods produced the same results for several different cubes. We assumed the others would correspond,also.



