Expert: Scotto - 10/31/2008

Question
12 wooden cubes, all different size and color. the largest cube is 12cm on a scale, the next is 11cm on a side and so on down to the smallest cube, which is 1cm on a scale. the cubes are colored red, blue, orange, yellow, green, purple, silver, white, violet, tangerine, crimeson, and mauve. the following relationship is hold among the cubes
1 the sum of the volume of the yellow and blue cubes equals the sum of the green and mauve cubes
2 the volume of thepurple cube alone equals the sum of the volumes of the red, violet, and silver cubes
3 the sum of the volumes of the blue, orange, purple, and violet cubes equals the sum of the volmnes of the mauve, green, tangerine, silver, white, and yellow cubes
4 the volumes of the green cube alone equals the sum of the volumes of the tangerine, purple, and yellow cubes

Match the size os each of the 123 cubes with its proper color


Answer
Lets label the volume of each of the colors red, blue, orange, yellow, green, purple, silver, white, violet, tangerine, crimeson, and mauve as R, B, O, Y, G, P, S, W, V, T, C, and M (that's neat that it's so easy to label them sinces they all start with a different letter).

(A) Y + B = G + M
(B) P = R + V + S
(C) B + O + P + V = M + G + T + S + W + Y
(D) G = T + P + Y

Note that the choices for volumes are the cubes from 1 to 12 which are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, and 1728.

Out of these, there is only one choice for equation (A).  As I remember reading somewhere, the smallest cubes that add up to the same result which have different values are 9^3+10^3=12^3+1^3.  This translates to 729 + 1000 = 1 + 1728 (could change order, switch sides).

For equation (B) and equation (D), we could have the values

216, 512, and 1000, who’s sum is 1728,
27, 64, and 125, who’s sum is 216, or
1, 216, and 512, who’s sum is 729.

Note that equation (A) and equation (B) contain entirely different values.  We know that out of these three choices for equation (B), we have the values 27, 64, 125 an 216.  This says that the value of P is 216.  Since the value of P is 216, from the remaining two out of three choices, we know that equation (D) has values 1, 216, or 512, and G has the value 729.

Since G has the value 729, M has the value 1000.  This means the others side of the equation has the values 1 and 1728.

Let’s review the choices
B 1 or 1728
C   
G 729
M 1000
O   
P 216
R   
S   
T   
V   
W   
Y 1 or 1728

That's not much, but at least it's three values that are assigned and 2 choices have been given to two other variables.

Let’s review the equations
(A) Y + B = 729 + 1000
(B) 216 = R + V + S
(C) B + O + 216 + V = 1000 + 729 + T + S + W + Y
(D) 729 = T + 216 + Y

Since equation (D) says that Y is 1 or 8 and equation (A) says that Y is 1 or 12, we know that Y is 1.

From equation (A) and Y having the value 1, this means that B has the value 12.  Again, since Y has the value 1, from equation (D), we know that T has the value 8.

Let’s review the choices once again
B 1728
C   
G 729
M 1000
O   
P 216
R   
S   
T 512
V   
W   
Y 1

and our equations are now

(A) 1 + 1728 = 729 + 1000 (done),
(B) 216 = R + V + S,
(C) 1728 + O + 216 + V = 1000 + 729 + 512 + S + W + 1, and
(D) 729 = 512 + 216 + 1 (done).

Equations (A) and equation (D) are done, so they won't be mentioned again.  This says that R, V, and S have the values 27, 64, and 125, but not necessarily in that order.

What that means is that C, O, and W are 8, 343, and 1331, but not necessarily in that order.

Now if we look at equation (C), we have 1,944+O+V = 2242+S+W, which goes to 298=O+V-S-W.

As has been said, the numbers we have left are 8, 27, 64, 125, 343, and 1331.

Since W is subtracted off in our rewritten (C) it has to be 8, since 343 and 1331 are too high.

This gives us the values
B 1728
C   
G 729
M 1000
O   
P 216
R   
S   
T 512
V   
W 8
Y 1

and our equations left that have variables in them are now

(B) 216 = R + V + S
(C) 1944 + O + V = 2250 + S.

The choices left for R, V, and S are (in whatever order) 27, 64, 125 and for C and O the values are 343 and 1331, in whatever order.

The values for O must be 343 since having a value of 1331 would make the left side of equation (C) too large, so C has the value 1331.

This gives
B 1728
C 1331
G 729
M 1000
O 343
P 216
R    
S   
T 512
V   
W 8
Y 1

and our equations are now

(B) 216 = R + V + S, and
(C) 1944 + 343 + V = 2250 + S, which is

(B) 216 = R + V + S, and
(C) 37 + V = S.

The values for the variables we haven’t used are 27, 64, and 125.  Equation (C) gives us that V must have the value 27 and S must have the value 64.  That leaves the value 125 for R.

My final answer is (from smallest to greatest)
Y 1
W 8
V 27
S 64
R 125
P 216
O 343
T 512
G 729
M 1000
C 1331
B 1728

Checking back at our first equations gives
(A) Y + B = G + M
(B) P = R + V + S
(C) B + O + P + V = M + G + T + S + W + Y
(D) G = T + P + Y,

which after inputting the values give us

(A) 1 + 1728 = 729 + 1000.
(B) 216 = 125 + 27 + 64.
(C) 1728 + 343 + 216 + 27 = 1000 + 729 + 512 + 64 + 8 + 1.
(D) 729 = 512 + 216 + 1, all of which are true.  Note that the value for equation (C) on both sides is 2314.

The length of a side on each of the blocks is then (in descending order, just to be different):
Black 12, Crimson 11, Mauve 10, Green 9, Tangerine 8, Orange 7,
Purple 6, Red 5, Silver 4, Violet 3, White 2, and Yellow 1.