
#1
Aug3004, 04:26 AM

P: 18

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?????? 



#2
Aug3004, 05:56 AM

P: 403

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



#3
Sep204, 04:17 AM

P: 18

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




#4
Sep204, 06:20 AM

Emeritus
Sci Advisor
PF Gold
P: 11,154

iq test if i have two tetrahedral pyramids
I hope that answer is not some giant number, and that there will be a solution within 1000...but I'm having doubts.




#5
Sep204, 08:41 AM

Sci Advisor
HW Helper
P: 2,004

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: [tex]\sum_{i=0}^ni^2=\frac{1}{6}n(n+1)(2n+1)[/tex] so the equation to solve is: [tex]2\frac{1}{6}n(n+1)(2n+1)+20=\frac{1}{6}(n+1)(n+2)(2n+2)[/tex] 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. 



#6
Sep204, 09:01 AM

Emeritus
Sci Advisor
PF Gold
P: 11,154

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 [tex]\sum_{i=0}^n{i(i+1)/2}=\frac{1}{12}n(n+1)(2n+1) + \frac {n(n+1)}{4}[/tex] 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. 



#7
Sep204, 09:41 AM

Sci Advisor
HW Helper
P: 2,004

I can' t count. 



#8
Sep204, 10:58 AM

P: 640

In white.
751966976 



#9
Sep204, 04:23 PM

P: 18

I´m afraid you're wrong.
Clue in white: The answer is below 1000 



#10
Sep204, 04:51 PM

P: 640

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. 


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