How Many Balls Are in the Base of a Triangular Pile of 120 Balls?

[KAISER91]

Homework Statement

To build a triangular pile consisting of 120 balls, the number of balls in the base would have to be??

Multiple Choice

a. 24
b. 30
c. 36
d. 45

Homework Equations

The Attempt at a Solution

The answer is 36.

How do I get this answer and is there a formula?


Thanks

 

[NobodySpecial]

Assuming the top layer has 1 ball, the second layer 3 (about the minimum you need for a triangle!) you should be able to work out how many the next layer has and so on

drawing it might help

 

[spyrustheviru]

I can't really understand the problem. When you talk about triangular pile, you mean something like this http://en.wikipedia.org/wiki/File:Números_triangulares.png?

because if it is this, then the answer should be 16. This is based on triangular numbers. let a(n) be the fuction that gives the n triangular number, a(n)=1+2+3+..+n. So the first TN is 1, the second is 3 and so on. If we think of those as number of dots, or balls in your case, starting from top to bottom, with the number of the n+1 row having one more ball than row n(and that number happens to be n), every one of those numbers is a total number of balls that can form a triangle. Check it yourself!
What you want is your triangular number to be 120(the sum of all the balls), so you look for n when a(n)=120. here http://oeis.org/A000217 we can see it is the 16th number, thus the bottom row has 16 balls. By hand validation, you get 16+15+14+..+1=120. On the same site you can see the formula for calculating the sum. Going a step further, the sum 1+2+3+..+(n-1)+n can be written (1+n)+(n-1+2)+...n/2 times...+(n+1)=n*(n+1)/2. This is equal to 120, and you solve with n as the unknown.

Someone give us some ideas, who is wrong?

 

[NobodySpecial]

I think a pile means in 3 dimensions - imagine a stack of oranges (although cannonballs always seems to be the standard example for some reason)

 

[Mark44]

I think a pile means in 3 dimensions - imagine a stack of oranges (although cannonballs always seems to be the standard example for some reason)

Yes, they mean in three dimensions.

 

[osnarf]

spirus, what you are looking at when you see it is the 16th number is that it is the 16th layer (15th if you don't count the layer that has zero balls). so in the 15th layer there would be 120 balls... all together there would be 0+1+3+6+...+105+120 balls which i don't feel like adding up but it would be a lot more than 120. You are keeping a running sum going as you go through that sequence and you want to find the layer that brings your sum up to 120.

 

[spyrustheviru]

spirus, what you are looking at when you see it is the 16th number is that it is the 16th layer (15th if you don't count the layer that has zero balls). so in the 15th layer there would be 120 balls... all together there would be 0+1+3+6+...+105+120 balls which i don't feel like adding up but it would be a lot more than 120. You are keeping a running sum going as you go through that sequence and you want to find the layer that brings your sum up to 120.

I'm sorry, I missunderstood the problem. I imagined this was on 2 dimentions, so my solutions in that case would be correct.
to the 3 dimentions now:
If I get it right, every layer is an even triangle, and thus it has 1,3,6,...x balls total, correct? Now, if you see these numbers, they are actually triagonal numbers...again!Our sum is 1+(1+2)+(1+2+3)+(1+2+3+4)+...+(1+2+3+4+..+k)=a(1)+a(2)+..+a(n), which is 120 in your problem.
I can't find a smart way to calculate the above sum. anyone got a formula ?

 

[eachus]

I didn't show the formula or how to get it, since this appeared to be a multiple choice question in a timed test. Once you realize that each layer must be a triangle number, and only one of the choices is a triangle number you are done.

But assume that the problem was to find the number of balls in the base for a large number of total balls. Now you have to drag out the heavy artillery. If we call the function which gives you the total number of balls given the length of a side T, we have:

T(1) = 1
T(n) = T(n-1) + n(n+1)/2

How do you solve the recurrence? In theory, you guess. But you have a lot of experience to go on. The answer will be a cubic equation in n, will have a divisor of 6, and will produce only whole numbers. The series we have to fit is 1,4,10,20,35...

(n)(n+1)(n+2)/6 = 1, 4, 10, 20...

Looks good to me. ;-)

So T(n) = (n)(n+1)(n+2)/6

For 120 balls,
120 = n(n+1)(n+2)/6
720 = n(n+1)(n+2) You can solve the cubic equation, reduce it to a quadratic, or notice that 8x9x10 = 720 and solve for n by inspection.

Substitute in n(n+1)/2 and get 36 as the correct answer for the original problem.