Tuesday, September 06, 2011

Math Beast Challenge problem

From MGRE's Math Beast Challenge page, we have the following problem:

This Week's Problem: "Frosted Cake"

Q: A cube-shaped cake is frosted on all sides except the bottom and then cut into 27 equally-sized cube-shaped pieces.

Quantity A
The number of pieces of cake that have exactly two sides frosted

Quantity B
One more than the number of pieces of cake that have fewer than two sides frosted


(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.

Have at it. My guess will be in the comments.



Kevin Kim said...

Draw the cake as a cube-shaped pile of blocks stacked 3 x 3 x 3. Imagine frosting on all sides of the cake except the bottom (very important).

How many small, cube-shaped slices have exactly two sides frosted?

Top layer of 9: 4 (mid-edge pieces)
Middle 9: 4 (the corner pieces)
Bottom 9: 4 (corner pieces again: remember that the bottom isn't frosted!)

So Quantity A equals 12. What about Quantity B?

"Fewer than 2 sides frosted" means "one or no sides frosted."

Top 9: 1 (center slice only)
Mid 9: 5 (mid-edge slices plus the very center slice, which is in too deep to be frosted)
Bottom 9: 5 (mid-edge slices, plus that central slice)

That's 11. Quantity B is one plus that number, i.e., 12.

So our answer is C: the quantities are equal.

My guess, anyway.

Surprises Aplenty said...

2 sides frosted = 12
more than 2 sides frosted=4
less than two sides frosted:
top down through middles =3
inside pieces, row 2 and 3 = 8
total = 11 (+1 = 12)

'C' looks correct. checksum: 12+4+11= 27 : okay!

Elisson said...

And the winner is... Kevin!

(This exercise gets trickier when you make a bigger cake, say one that is 12 x 12 x 12 slices. Or n x n x n slices, in which case one must derive a formula using Schroedinger's wave equation and the speed of light in a Hoover vacuum.)

Horace Jeffery Hodges said...

I insist the answer is "D" . . . for me, anyway.

Jeffery Hodges

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