Can you fill in a Rubik's Cube with numbers from 1 to 27 so that opposite faces have equal sum?

Take a Rubik's Cube and you can see that it is made up of 27 unit cubes out of which 26 are directly visible to our eyes, say outer unit cubes.

19 unit cubes are visible from this view. Count !

Here is the task for you:

Tear a piece of paper into 27 smaller pieces (in the size of a face of smaller unit cube) and write down the numbers from 1 to 27 one on each of it. Now, paste these pieces (except some one piece), exactly one on each of the 26 outer unit cubes in such a way that the sum of numbers on the unit cubes in any two opposite faces are equal.

Stop here and try out the task yourself for at least an hour and then scroll down for more ideas on it.......


Convert the problem into 2D, i.e., first ask yourself whether you can fill a 3\times 3 grid with numbers from 1 to 9 such that sum of numbers on opposite sides have equal sum. For example, given below are two such grids

You may as well consider 3\times 3 magic squares in which the sum of numbers in each of the rows, columns and the two main diagonals are all equal as shown below

Note that there multiple examples for these kind of 3\times 3 grids and they can be obtained systematically by assigning variables to each unit square and considering few initial assumptions. Try out yourself and provide more examples of such grids in the comments below !

Now use anyone of these 3\times 3 grids to construct cleverly a 3\times 3\times 3 cube so that sum of numbers on the unit cubes in any two opposite faces are equal. Try for sometime and scroll down for more ideas on it.......


The numbers from 1 to 27 can be written as

\{9(0)+1, 9(0)+2, \cdots, 9(0)+9, 9(1)+1, 9(1)+2, 9(1)+3, \cdots, 9(1)+9, 9(2)+1, 9(2)+2, \cdots, 9(2)+9\}

Consider the 3 layers of the cube from bottom to top as 3\times 3 grids as shown below that is filled similar to the 3\times 3 grid above except for the values of k_i, k_i', k_i''.

Now notice that \{k_i,k_i',k_i''\}=\{0,1,2\} in some order. So, the two pairs of opposite faces formed by (R_1, R_3)\ ;\ (C_1, C_3) of the 3\times 3 grids have equal sum by virtue of its construction. More importantly, the sum of first grid should equal sum of the last grid because they are the 3^{rd} pair of opposite faces. For this to happen, we evenly distribute the values of k's by taking

(k_1,k_2,\cdots,k_9)=(0,1,2,0,\cdots,2)\ ;\ (k_1',k_2',\cdots,k_9')=(1,2,0,1\cdots,0)\ ;\ (k_1'',k_2'',\cdots,k_9'')=(2,0,1,2,\cdots,1)

and hence the following distribution of numbers in the 3 layers of cube make each of the 3 pairs of opposite faces to have equal sum.

Use this to configuration to fill in the Rubik's Cube and check out yourself. Do similar things for other 3\times 3 grids at the start and comment out few other possible Rubik's Cube configurations !

At sumonmath.com program, we have one activity problem like this in each worksheet for every fortnight (Bi-weekly). We are really enthusiastic about the growth of a kid who solves non-routine problems like this every fortnight !

Join us at sumonmath.com to solve more interesting problems like this. Don't worry, we provide hints and detailed solution to every problem of the worksheet as hard copy to your home !

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