Math mini-contest problem for Day 26 on D.Buzz for February 2021 😎

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Math mini-contest problem for Day 26 on D.Buzz for February 2021 😎

Fang bought a rectangular cake with icing on all sides except the bottom. He then cut the cake into 6 by 8 by 5 cubes. How many of the cake cubes will have at least 2 sides with icing?

math mathematics minicontest dbuzz cake cube
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(Edited)

40

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I'm assuming the cube counts are given in the conventional x,y,z order. It matters if z var changes. Count along x twice, then the along y twice subtracting two each time for corners already counted then add 4 times z again minus the corner 4.
2x + 2(y - 2) + 4(z - 1)

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The order matters, although x and y are interchangeable, z is not. You can see in my working.


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My guess is 48 slices with at least 2 sides of icing. Work is in the next comment.

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This can be found by using the perimeter of the cake minus the perimeter of the side with no icing. The reason why I used the perimeter is because this where the icing is.

4(x+y+z) - 2(x+y)
4(6+8+5) - 2(6+8) = 76-28=48.

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I think you're over counting the corners.

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I might be but I am too lazy to check. I kind of just based on it on easy way to calculate. It's late at night.

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No solution.... We don't know the dimensions of the cake

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That's what I first thought too, but it doesn't really matter because the cake is cut into cubes. Since the cubes have equal sides you can assume the unit of measure is in cube-side-lengths if it helps.

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Hmm... But imagine if the cakes height was 2ce the height of the cube.... U can cut a cake horizontally too ... Right?

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Yes, it's cubes. Maybe think of each cube as having sides of 1cm if it helps. Once you work it out I think you'll see why the units don't actually matter.

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assuming 6 and 8 to be x and y and 5 to be z/height: 60
assuming 8 to be z: 54
assuming 6 to be z: 56
(out of competition, just because I like it :) )

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(Edited)

Answer for Day 26 Math Problem

  • 40 cubes 🎯

  • 42 cubes 🎯

  • 46 cubes 🎯

Any of the three answers shall be accepted, due to possible differing solvers' perspectives regarding the original problem. No solution required (but is recommended)!

Solutions

First, we should take note that we can look at the cake in 5 different positions (top, left, right, front, and back). Looking at either the left or right, or front and back, will give the same results.

If the solver assumes that the dimensions 6 by 8 by 5 (cubic units) are in the conventional xyz order and facing the xy plane, then there are 6 cubes horizontally, 5 cubes vertically, and 5 cubes forward. We will simply need to count all the cubes which have 2 or 3 sides with icing on the edges and corners.

This solution will assume that the cake is 6 cubes wide, 8 cubes high, and 5 cubes long.

  • For each layer at the bottom (1st) layer to the 7th layer, there are 4 qualifying cubes, which are at the edges. All of them have 2 sides with icing. There are a total of 28 qualifying cubes.
  • On the top (8th) layer, all the edges and corners qualify. There are 4 corner cubes (whose 3 sides have icing) and 14 edge cubes (whose 2 sides have icing), for a total of 18 cubes.

There are a total of 46 qualifying cubes using the conventional ordering of dimensions and plane of reference.

If you interchange the width (x) and the height (y) to 8 cubes wide and 6 cubes high (and length is still 5 cubes long), you will get the following:

  • For each layer at the 1st layer to the 5th layer, there are 4 qualifying cubes, which are at the edges. All of them have 2 sides with icing. There are a total of 20 qualifying cubes.
  • On the top (6th) layer, all the edges and corners qualify. There are 4 corner cubes (whose 3 sides have icing) and 18 edge cubes (whose 2 sides have icing), for a total of 22 cubes.

There are a total of 42 qualifying cubes when the width and height are interchanged.

If the height and length are interchanged (such that the dimensions are now 6 by 5 by 8 cubic units), we will get the following:

  • For each layer at the 1st layer to the 4th layer, there are 4 qualifying cubes, which are at the edges. All of them have 2 sides with icing. There are a total of 16 qualifying cubes.
  • On the top (5th) layer, all the edges and corners qualify. There are 4 corner cubes (whose 3 sides have icing) and 20 edge cubes (whose 2 sides have icing), for a total of 24 cubes.

There are a total of 40 qualifying cubes when the width and length are interchanged.

If the width and length are interchanged (such that the dimensions are now 5 by 8 by 6 cubic units), we will get 28 qualifying cubes from the first 7 layers and 18 from the top layer, giving a total of 46 qualifying cubes.

If the width becomes the height, the height becomes the length, and the length becomes the width (such that the dimensions are now 8 by 5 by 6 cubic units), we will get 16 qualifying cubes from the first 4 layers and 24 from the top layer, giving a total of 42 qualifying cubes.

Winner: @eturnerx-dbuzz 🏅

1 HIVE has been sent to @eturnerx-dbuzz's Hive account. 💰

Mentions: @jfang003, @holovision, @ahmadmanga (@ahmadmangazap), and @minus-pi 🤔
 

Please tell me if you think there is something wrong with these solutions. Thanks! 😅

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On the top (8th) layer, all the edges and corners qualify. There are 4 corner cubes (whose 3 sides have icing) and 10 edge cubes (whose 3 sides have icing), for a total of 14 cubes.

The edges on 6 squares x 5 squares are: The 4 corners, plus counting the edges (not corners) clockwise from a long edge, 4 + 3 + 4 + 3 = 14, so 18. Adding the 28 edges on the 7 vertical layers, gives 46.

Intuitively, the length on the vertical only "loses" one to a corner but is multiplied by four. So, when the largest length is on the vertical it should give the biggest result.

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Thanks for correcting my solution! 😅 It sometimes becomes hard for me to publish the solution within 24 hours after the problem is published (through schedule), so I make minor mistakes especially on longer solutions. 🤯😅

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