Math Question for Fun!

[Debater Slayer]

Suppose you have this two-dimensional pyramid of cups:

How can you count the number of cups without counting the cups individually?

Bonus question: Assume this pyramid is turned into a pyramid with four faces, with each face looking like the pyramid in the picture. How many cups are there in that tetrahedral pyramid?

Please use spoiler tags for your answers!

 

[Jedster]

Spoiler: look

∑ N(N+1)/2 from N=1 to 12 yields 78

 

[Debater Slayer]

Spoiler: look

∑ N(N+1)/2 from N=1 to 12 yields 78

Spoiler

I can't post the answer yet, but what about the bonus question?

 

[Subduction Zone]

Spoiler

"Get my OCD cousin to count them for me. Bonus question, the same. It would take her a bit longer, but it would give her a great sense of accomplishment. Wins all around!"]

 

[anna.]

I hate math and haven't the slightest idea what the answer is but wanted to bump your thread.

 

[Jedster]

Spoiler

I can't post the answer yet, but what about the bonus question?

Spoiler: bonus

that is 4 faces plus the base which is 12X12 ETA(assuming the pyramid is not solid or is it?)

i.e. 4x78 + 144 =456

 

[Subduction Zone]

Spoiler: bonus

that is 4 faces plus the base which is 12X12

i.e. 4x78 + 144 =456

I think that as written it would be a solid pyramid. Not just the faces. My cousin is still working on it. You have to add a series of squares, or wait for my cousin. She is still counting.

 

[Jedster]

Suppose you have this one-dimensional pyramid of cups:

How can you count the number of cups without counting the cups individually?

Bonus question: Assume this pyramid is turned into a pyramid with four faces, with each face looking like the pyramid in the picture. How many cups are there in that tetrahedral pyramid?

Please use spoiler tags for your answers!

Spoiler: solid pyramid

If the pyramid is solid full of cups then the total is

∑n.n from n=12 to n=1
using Σn2 = [n(n+1)(2n+1)]/6
12.13.25/6= 650

 

[Subduction Zone]

Spoiler: solid pyramid

If the pyramid is solid full of cups the the total is

∑n.n from n=12 to n=1
using Σn2 = [n(n+1)(2n+1)]/6
12.13.25/6= 650

My cousin agrees

Sum of n, n², or n³ | Brilliant Math & Science Wiki

 

[Debater Slayer]

I hate math and haven't the slightest idea what the answer is but wanted to bump your thread.

Thank you! I appreciate it.

 

[Debater Slayer]

Spoiler: bonus

that is 4 faces plus the base which is 12X12 ETA(assuming the pyramid is not solid or is it?)

i.e. 4x78 + 144 =456

Spoiler

Not solid, no, but the way I have in mind for counting the cups making up the four faces would still different from the above. I'll share later!

 

[fantome profane]

Spoiler

1+2+3+4+5+6+7+8+9+10+11+12 = 78 for the 2 dimensional. I was trying to remember if there is a word for that (like factorial, but not factorial) but if there is I can't remember it.

 

[Debater Slayer]

Spoiler

1+2+3+4+5+6+7+8+9+10+11+12 = 78 for the 2 dimensional. I was trying to remember if there is a word for that (like factorial, but not factorial) but if there is I can't remember it.

And 78 X 12 for the 3 dimensional pyramid.

Spoiler

It's called an arithmetic series.

 

[Jedster]

My cousin agrees

Sum of n, n², or n³ | Brilliant Math & Science Wiki

Spoiler

1+2+3+4+5+6+7+8+9+10+11+12 = 78 for the 2 dimensional. I was trying to remember if there is a word for that (like factorial, but not factorial) but if there is I can't remember it.

Spoiler: alternatively

Notice that 1+12 =13, 2+11=13 etc , so you have 6 pairs that each add up to 13

So , if one is inclined , one could easily add up the first 100 integers or even 1000

 

[LuisDantas]

Spoiler

I think you mean a two-dimensional pyramid, right?

This is an use of the concept of triangular numbers, for fairly obvious reasons.

Triangular number - Wikipedia

en.wikipedia.org

We clearly have N rows, each with the number of cups of each natural number up to N (inclusively).

That means that, taking the earliest and latests rows together in pairs, we have N/2 x (N + 1) cups if N is even, or (N-1)/2 * (N+1) + N cups if N is odd.

In this case, N is 12, therefore the formula translates into 6 x 13, which means 78 cups in all.


Now, for the three dimensional, four-sided pyramid, we would have twelve levels counting downwards from one to 144 (12 * 12) cups. The higher level has a sole cup, the next higher has four, then 9, 16, ..., 144.

That by its turn amounts to the sum of 12 numbers. Let's describe them as an array C[1..12], where C[1] = 1, C[2] = 4, C[3] = 9, etc, up to C[12] = 144.

Each C[n] is equivalent to n squared, of course. But also to C[n-1] + n + (n - 1). I believe there is a technique called an "evanescent triangle" that helps in detecting this pattern, which I read about long ago.

There is probably some way of making a simpler calculation of the sum, but I fail to conceive of it right now.

 

[Debater Slayer]

I may post the answer soon! Tagging @exchemist because I think he might enjoy this.

 

[ChristineM]

Spoiler

Easy way.

It's a pyramid so the cups follow a progression of x+1 for each row

1+2+3...+12 = 78

For the 3D pyramid
It's front + back 78+78 = 156
Plus 2 sides 1+2+3+10 (note the end cups are already counted) =55 *2 =110
Total cups 266

Unless it's a solid pyramid then ???

I hope!

 

[exchemist]

I may post the answer soon! Tagging @exchemist because I think he might enjoy this.

I'm still trying to work out why it is not 72, i.e. half of 12 x 12.

 

[Subduction Zone]

I'm still trying to work out why it is not 72, i.e. half of 12 x 12.

That is what one gets by making a rhombus of cups, but that is another way to solve it.

 

[Debater Slayer]

I'm still trying to work out why it is not 72, i.e. half of 12 x 12.

I can explain why if you want. That's a great and potentially confusing question, though.