Iq test if i have two tetrahedral pyramids

[datatec]

this one i got from an iq test

if i have two tetrahedral pyramids of equal sizes made of golf balls and I combine them (presuming every ball is used) the minumum amount of balls needed would be 20 to make one larger pyramid. What would the minimum be should the two smaller pyramids be different sizes?

 

[Rogerio]

Select to see...:-)

The same 20 balls (you may have 2 pyramids made of 4 and 10 balls).

 

[datatec]

Nope

You need to use all the balls. In your case you have six remaining.

 

[Gokul43201]

I hope that answer is not some giant number, and that there will be a solution within 1000...but I'm having doubts.

 

[Galileo]

I don't think I understand the question correctly.
You said if you combine the number of balls from two equally sized tetrahedral pyramids and add 20, then you have the number of balls needed to make a pyramid one size larger.
The number of balls needed for a pyramid of size n is:
$$\sum_{i=0}^ni^2=\frac{1}{6}n(n+1)(2n+1)$$
so the equation to solve is:
$$2\frac{1}{6}n(n+1)(2n+1)+20=\frac{1}{6}(n+1)(n+2)(2n+2)$$
to find the size of the smaller pyramids.
But it has no integer solutions...

Possibly, my picture of making pyramids with golfballs is wrong.
Stacking them like this will not make them tetrahedral, but I don't see any other way to do it.

 

[Gokul43201]

I don't think I understand the question correctly.
You said if you combine the number of balls from two equally sized tetrahedral pyramids and add 20, then you have the number of balls needed to make a pyramid one size larger.
The number of balls needed for a pyramid of size n is:
$$\sum_{i=0}^ni^2=\frac{1}{6}n(n+1)(2n+1)$$
so the equation to solve is:
$$2\frac{1}{6}n(n+1)(2n+1)+20=\frac{1}{6}(n+1)(n+2)(2n+2)$$
to find the size of the smaller pyramids.
But it has no integer solutions...

Possibly, my picture of making pyramids with golfballs is wrong.
Stacking them like this will not make them tetrahedral, but I don't see any other way to do it.

The question is simply, "What 2 dissimilar tetrahdra (built from balls) can be combined to make a new tetrahedron, using exactly the number of balls contained in the smaller ones ?"

The case of 20 balls would be a solution were this question lacking the word 'dissimilar'. It has nothing to do with the current question.

Next, a tetrahedral number of height n, is the sum of the first n triangular numbers, and so, should be given by the sum

$$\sum_{i=0}^n{i(i+1)/2}=\frac{1}{12}n(n+1)(2n+1) + \frac {n(n+1)}{4}$$

Triangular numbers are : 1,3,6,10,15,21,28,...
Tetrahedral numbers are : 1,4,10,20,35,56,84,120,...

We are looking for 2 different Tet. Numbers that add to give a third one.

 

[Galileo]

Next, a tetrahedral number of height n, is the sum of the first n triangular numbers, and so, should be given by the sum

Oh yeah... a tetrahedron has 4 sides.
I can' t count.

 

[K.J.Healey]

In white.
751966976

 

[datatec]

Wrong Again

I´m afraid you're wrong.
Clue in white:

The answer is below 1000

 

[K.J.Healey]

Ahh, i was doing half of an octahedron, not a tetrahedron.

ANSWER :
680 balls, by a height of 14 and a height of 8 TH's to make a height of 15 TH.