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:

(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.


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4 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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